{"id":752583,"date":"2025-01-09T17:25:19","date_gmt":"2025-01-09T11:55:19","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=752583"},"modified":"2025-01-09T17:25:19","modified_gmt":"2025-01-09T11:55:19","slug":"electrostatic-potential-and-capacitance-class-12-mcq-with-answers","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/mcqs\/class-12-physics-electrostatic-potential-and-capacitance\/","title":{"rendered":"Electrostatic Potential and Capacitance Class 12 MCQ with Answers"},"content":{"rendered":"

Electrostatics is a fundamental topic in physics that deals with charges at rest and the forces, fields, and potentials associated with them. One of the key areas of study in electrostatics is the concept of electrostatic potential<\/strong> and capacitance<\/strong>, which are essential for understanding the behavior of electric fields and energy storage in capacitors. These topics are not only vital for academic purposes but also have numerous real-world applications in electronic devices, energy systems, and even in medical technology.<\/p>\n

In the Class 12 Physics curriculum, “Electrostatic Potential and Capacitance” is a significant chapter that lays the foundation for advanced studies in electromagnetism and electrical engineering. Mastering this chapter requires a clear understanding of the theoretical concepts, mathematical derivations, and their practical implications. Multiple Choice Questions (MCQs)<\/a><\/strong> play a crucial role in testing and reinforcing this understanding, as they challenge students to think critically and apply their knowledge in various scenarios.<\/p>\n

This chapter introduces students to electrostatic potential<\/strong>, which is the work done in moving a unit positive charge from infinity to a given point in an electric field without acceleration. It helps in visualizing how charges interact and how energy is distributed in an electric field. Concepts like potential difference<\/strong> and equipotential surfaces<\/strong> are pivotal in understanding the nature of electric fields in different configurations.<\/p>\n

The concept of capacitance<\/strong>, on the other hand, explores how electrical energy is stored in systems known as capacitors. A capacitor is a simple yet powerful device used in countless electronic applications, from filtering signals in circuits to storing energy in power systems. The study of capacitors involves understanding parameters like the relationship between charge, voltage, and capacitance, the effect of dielectric materials, and the behavior of capacitors in series and parallel combinations.<\/p>\n

Practicing MCQs helps students identify areas where they might need further clarity while also preparing them for competitive exams like JEE<\/a><\/strong>, NEET<\/a><\/strong>, and other entrance tests, where such questions are commonly featured. These questions often test conceptual clarity, problem-solving skills, and the ability to think critically under time constraints.<\/p>\n

This compilation of Class 12 MCQs on “Electrostatic Potential and Capacitance” includes detailed answers and explanations to aid students in self-assessment and learning. Each question has been carefully selected to represent different levels of difficulty, ensuring that students are exposed to a variety of problem types. Whether you\u2019re preparing for board exams, competitive tests, or simply looking to deepen your understanding of physics, these MCQs serve as an excellent resource for comprehensive learning.<\/p>\n

Class 12 Electrostatic Potential and Capacitance MCQ with Answers<\/h2>\n

Question: A hollow metal sphere with a radius of 5 cm is charged so that the potential on its surface is 10 V. What is the potential at its center?<\/strong><\/p>\n

(a) 0 V
\n(b) 10 V
\n(c) Same as at a point 5 cm away from the surface
\n(d) Same as at a point 25 cm away from the surface<\/p>\n

Answer<\/strong>: (b) 10 V<\/p>\n

Explanation<\/strong>: Inside a hollow metal sphere, the electric potential is uniform and equal to the potential on the surface. Since the surface potential is 10 V, the center of the sphere also has a potential of 10 V.<\/p>\n

Question: What happens when a positively charged particle is released from rest in a uniform electric field?<\/strong><\/p>\n

(a) Its electric potential energy stays the same because the field is uniform.
\n(b) Its electric potential energy increases as it moves along the field.
\n(c) Its electric potential energy decreases as it moves along the field.
\n(d) Its electric potential energy decreases as it moves opposite to the field.<\/p>\n

Answer<\/strong>: (c) Its electric potential energy decreases as it moves along the field.<\/p>\n

Explanation<\/strong>: A positively charged particle moves in the direction of the electric field due to the force acting on it. As it moves, its electric potential energy decreases because work is being done by the field.<\/p>\n

Question: Equipotential surfaces at a large distance from a group of charges with a net charge are approximately:<\/strong><\/p>\n

(a) Spheres
\n(b) Planes
\n(c) Paraboloids
\n(d) Ellipsoids<\/p>\n

Answer<\/strong>: (a) Spheres<\/p>\n

Explanation<\/strong>: From a large distance, a group of charges appears like a single point charge. For such charges, the potential is the same at all points equidistant from the center, forming spherical equipotential surfaces.<\/p>\n

Also Check: JEE Main Physics Electrostatics Previous Year Questions with Solutions<\/a><\/strong><\/em><\/p>\n

Question: Why is water not used as a dielectric between the plates of a capacitor?<\/strong><\/p>\n

(a) Its dielectric constant is very low.
\n(b) Its dielectric strength is very low.
\n(c) Its dielectric constant is very high.
\n(d) Its dielectric strength is very high.<\/p>\n

Answer<\/strong>: (b) Its dielectric strength is very low.<\/p>\n

Explanation<\/strong>: While water has a high dielectric constant, its dielectric strength (the ability to withstand high electric fields without breaking down) is very low. This makes it unsuitable as a dielectric material in capacitors.<\/p>\n

Question: If a unit positive charge is moved along an equipotential surface, what happens to the work done?<\/strong><\/p>\n

(a) Work is done on the charge.
\n(b) Work is done by the charge.
\n(c) Work done is constant.
\n(d) No work is done.<\/p>\n

Answer<\/strong>: (d) No work is done.<\/p>\n

Explanation<\/strong>: On an equipotential surface, the electric potential is the same at all points. Moving a charge along such a surface requires no work because there is no change in potential energy.<\/p>\n

Question: The electric potential <\/strong><\/p>\nV<\/mi><\/mrow>V<\/annotation><\/semantics><\/math>\n

V<\/span><\/span><\/span> at any point<\/p>\nO<\/mi>(<\/mo>x<\/mi>,<\/mo>y<\/mi>,<\/mo>z<\/mi>)<\/mo><\/mrow>O(x, y, z)<\/annotation><\/semantics><\/math>\n

in space is given as<\/p>\nV<\/mi>=<\/mo>4<\/mn>x<\/mi>2<\/mn><\/msup><\/mrow>V = 4x^2<\/annotation><\/semantics><\/math>\n

volts. What is the electric field at point<\/p>\n(<\/mo>1<\/mn>\u2009<\/mtext>m<\/mtext>,<\/mo>0<\/mn>,<\/mo>2<\/mn>\u2009<\/mtext>m<\/mtext>)<\/mo><\/mrow>(1 \\, \\text{m}, 0, 2 \\, \\text{m})<\/annotation><\/semantics><\/math>\n

?<\/p>\n

(a) 8 along the negative x-axis
\n(b) 8 along the positive x-axis
\n(c) 16 along the negative x-axis
\n(d) 16 along the positive z-axis<\/p>\n

Answer<\/strong>: (a) 8 along the negative x-axis<\/p>\n

Explanation<\/strong>: The electric field is the negative gradient of potential:<\/p>\nE<\/mi>=<\/mo>\u2212<\/mo>d<\/mi>V<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac><\/mrow>E = – \\frac{dV}{dx}<\/annotation><\/semantics><\/math>\n

\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>.
\nGiven<\/p>\nV<\/mi>=<\/mo>4<\/mn>x<\/mi>2<\/mn><\/msup><\/mrow>V = 4x^2<\/annotation><\/semantics><\/math>\n

, we calculate<\/p>\nE<\/mi>=<\/mo>\u2212<\/mo>d<\/mi>(<\/mo>4<\/mn>x<\/mi>2<\/mn><\/msup>)<\/mo><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>=<\/mo>\u2212<\/mo>8<\/mn>x<\/mi><\/mrow>E = – \\frac{d(4x^2)}{dx} = -8x<\/annotation><\/semantics><\/math>\n

At<\/p>\nx<\/mi>=<\/mo>1<\/mn><\/mrow>x = 1<\/annotation><\/semantics><\/math>\n

,<\/p>\nE<\/mi>=<\/mo>\u2212<\/mo>8<\/mn><\/mrow>E = -8<\/annotation><\/semantics><\/math>\n

V\/m, directed along the negative x-axis.<\/p>\n

Question: A conductor with a positive charge:<\/strong><\/p>\n

(a) Always has a positive potential.
\n(b) Always has zero potential.
\n(c) Always has a negative potential.
\n(d) Can have positive, zero, or negative potential.<\/p>\n

Answer<\/strong>: (d) Can have positive, zero, or negative potential.<\/p>\n

Explanation<\/strong>: The potential of a charged conductor depends on the reference point for zero potential. A positively charged conductor can have any potential relative to the chosen reference.<\/p>\n

Question: A charged capacitor is disconnected from a battery and connected in parallel to an identical uncharged capacitor. What happens to the total electrostatic energy?<\/strong><\/p>\n

(a) Increases by a factor of 4.
\n(b) Decreases by a factor of 2.
\n(c) Remains the same.
\n(d) Increases by a factor of 2.<\/p>\n

Answer<\/strong>: (b) Decreases by a factor of 2.<\/p>\n

Explanation<\/strong>: When the charged capacitor shares its charge with the uncharged capacitor, the potential difference is halved. The energy is proportional to<\/p>\nV<\/mi>2<\/mn><\/msup><\/mrow>V^2<\/annotation><\/semantics><\/math>\n

, so the total energy decreases by a factor of 2.<\/p>\n

Question: Which of the following is false for a perfect conductor?<\/strong><\/p>\n

(a) The surface is an equipotential surface.
\n(b) The electric field outside the surface is perpendicular to it.
\n(c) The charge on the conductor is uniformly distributed on its surface.
\n(d) None of these.<\/p>\n

Answer<\/strong>: (d) None of these.<\/p>\n

Explanation<\/strong>: All the listed statements are true for a perfect conductor. It has equipotential surfaces, charges are uniformly distributed, and the electric field outside is perpendicular to the surface.<\/p>\n

Question: 64 small drops, each with capacitance <\/strong><\/p>\nC<\/mi><\/mrow>C<\/annotation><\/semantics><\/math>\n

and charge<\/p>\nq<\/mi><\/mrow>q<\/annotation><\/semantics><\/math>\n

, combine to form one large drop. What is the charge on the big drop?<\/p>\n

(a)<\/p>\n2<\/mn>q<\/mi><\/mrow>2q<\/annotation><\/semantics><\/math>\n

(b)<\/p>\n4<\/mn>q<\/mi><\/mrow>4q<\/annotation><\/semantics><\/math>\n

(c)<\/p>\n16<\/mn>q<\/mi><\/mrow>16q<\/annotation><\/semantics><\/math>\n

(d)<\/p>\n64<\/mn>q<\/mi><\/mrow>64q<\/annotation><\/semantics><\/math>\n

 <\/p>\n

Answer<\/strong>: (d)<\/p>\n64<\/mn>q<\/mi><\/mrow>64q<\/annotation><\/semantics><\/math>\n

 <\/p>\n

Explanation<\/strong>: The total charge is the sum of charges on all 64 small drops. Since charge is additive, the large drop holds<\/p>\n64<\/mn>q<\/mi><\/mrow>64q<\/annotation><\/semantics><\/math>\n

units of charge.<\/p>\n

Also Check: Class 12 Electrostatic Potential and Capacitance MCQs<\/a><\/strong><\/em><\/p>\n

Question: What is the work done by an external agent to move a unit positive charge from a point at +500 V to another point at \u2013500 V?<\/strong><\/p>\n

(a) +1000 J<\/p>\n

(b) \u20131000 J<\/p>\n

(c) +500 J<\/p>\n

(d) \u2013500 J<\/p>\n

Answer: <\/strong>\u20131000 J<\/strong><\/p>\n

Explanation:<\/strong>
\nThe work done in moving a charge is given by:<\/p>\n

 <\/p>\nW<\/mi>=<\/mo>q<\/mi>\u00d7<\/mo>(<\/mo>V<\/mi>f<\/mi><\/msub>\u2212<\/mo>V<\/mi>i<\/mi><\/msub>)<\/mo><\/mrow><\/semantics><\/math>\n

 <\/p>\n

For a unit positive charge (<\/p>\nq<\/mi>=<\/mo>+<\/mo>1<\/mn><\/mrow>q = +1<\/annotation><\/semantics><\/math>\n

), the initial potential (<\/p>\nV<\/mi>i<\/mi><\/msub><\/mrow>V_i<\/annotation><\/semantics><\/math>\n

V<\/span><\/span><\/span><\/span>) is +500 V, and the final potential (<\/p>\nV<\/mi>f<\/mi><\/msub><\/mrow>V_f<\/annotation><\/semantics><\/math>\n

\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span>) is \u2013500 V.<\/p>\n

 <\/p>\nW<\/mi>=<\/mo>1<\/mn>\u00d7<\/mo>(<\/mo>\u2212<\/mo>500<\/mn>\u2212<\/mo>500<\/mn>)<\/mo>=<\/mo>\u2212<\/mo>1000<\/mn>\u2009<\/mtext>J<\/mtext><\/mrow><\/semantics><\/math>\n

 <\/p>\n

Thus, the work done is \u20131000 J, which means the energy is released.<\/p>\n

Question: What happens to the electric potential energy when a positively charged particle moves along the direction of a uniform electric field?<\/strong><\/p>\n

(a) Remains constant because the field is uniform.<\/p>\n

(b) Increases because the charge moves along the field.<\/p>\n

(c) Decreases because the charge moves along the field.<\/p>\n

(d) Decreases because the charge moves opposite to the field.<\/p>\n

Answer: <\/strong>Decreases because the charge moves along the field.<\/strong><\/p>\n

Explanation:<\/strong>
\nWhen a positive charge moves along the electric field, it naturally moves towards a lower potential. As it does so, its potential energy decreases, and its kinetic energy increases due to acceleration by the field.<\/p>\n

Question: At what angle do electric field lines intersect equipotential surfaces?<\/strong><\/p>\n

(a) 0\u00b0<\/p>\n

(b) 180\u00b0<\/p>\n

(c) 90\u00b0<\/p>\n

(d) 270\u00b0<\/p>\n

Answer: <\/strong>90\u00b0<\/strong><\/p>\n

Explanation:<\/strong>
\nEquipotential surfaces are always perpendicular to electric field lines. This is because there is no work done in moving a charge along an equipotential surface, and the field direction must be orthogonal to maintain this condition.<\/p>\n

Question: What is the potential difference between two points if 40 J of work is required to move a charge of \u2013500 C from one point to another?<\/strong><\/p>\n

(a) 0.08 V<\/p>\n

(b) 0.10 V<\/p>\n

(c) 0.05 V<\/p>\n

(d) 0.02 V<\/p>\n

Answer:<\/strong> 0.08 V<\/strong><\/p>\n

Explanation:<\/strong>
\nThe potential difference (<\/p>\nV<\/mi><\/mrow>V<\/annotation><\/semantics><\/math>\n

) is calculated as:<\/p>\n

 <\/p>\nV<\/mi>=<\/mo>W<\/mi>q\u200b<\/span><\/mi><\/mfrac><\/mrow><\/semantics><\/math>\n

 <\/p>\n

Here,<\/p>\nW<\/mi>=<\/mo>40<\/mn>\u2009<\/mtext>J<\/mtext><\/mrow>W = 40 \\, \\text{J}<\/annotation><\/semantics><\/math>\n

and<\/p>\nq<\/mi>=<\/mo>\u2212<\/mo>500<\/mn>\u2009<\/mtext>C<\/mtext><\/mrow>q = -500 \\, \\text{C}<\/annotation><\/semantics><\/math>\n

 <\/p>\n

 <\/p>\nV<\/mi>=<\/mo>40<\/mn>500<\/mn><\/mfrac>=<\/mo>0.08<\/mn>\u2009<\/mtext>V<\/mtext><\/mrow><\/semantics><\/math>\n

 <\/p>\n

Also Check: Important Questions for CBSE Class 12 Physics Electrostatic Potential<\/a><\/strong><\/em><\/p>\n

Question: What happens to the capacitance of a parallel plate capacitor if the distance between the plates is tripled and the plate area is doubled?<\/strong><\/p>\n

(a) 10 \u03bcF<\/p>\n

(b) 40 \u03bcF<\/p>\n

(c) 80 \u03bcF<\/p>\n

(d) 100 \u03bcF<\/p>\n

Answer: <\/strong>40 \u03bcF<\/strong><\/p>\n

Explanation:<\/strong>
\nThe formula for capacitance is:<\/p>\n

 <\/p>\nC<\/mi>=<\/mo>\u03b5<\/mi>0<\/mn><\/msub>A<\/mi><\/mrow>d\u200b<\/span><\/mi><\/mfrac><\/mrow><\/semantics><\/math>\n

 <\/p>\n

If the area (<\/p>\nA<\/mi><\/mrow>A<\/annotation><\/semantics><\/math>\n

) is doubled, and the distance (<\/p>\nd<\/mi><\/mrow>d<\/annotation><\/semantics><\/math>\n

) is tripled, the capacitance becomes:<\/p>\n

 <\/p>\nC<\/mi>new<\/mtext><\/msub>=<\/mo>C<\/mi>\u00d7<\/mo>2<\/mn>3\u200b<\/span><\/mn><\/mfrac><\/mrow><\/semantics><\/math>\n

 <\/p>\n

For an initial capacitance of 60 \u03bcF, the new capacitance is:<\/p>\n

 <\/p>\nC<\/mi>new<\/mtext><\/msub>=<\/mo>60<\/mn>\u00d7<\/mo>2<\/mn>3<\/mn><\/mfrac>=<\/mo>40<\/mn>\u2009<\/mtext>\u03bc<\/mi>F<\/mtext><\/mrow><\/semantics><\/math>\n

 <\/p>\n

Question: If 27 small water drops, each charged to 220 V, combine to form one larger drop, what is the potential of the larger drop?<\/strong><\/p>\n

(a) 1320 V<\/p>\n

(b) 1520 V<\/p>\n

(c) 1980 V<\/p>\n

(d) 660 V<\/p>\n

Answer: <\/strong>1980 V<\/strong><\/p>\n

Explanation:<\/strong>
\nWhen small drops combine, the volume and charge add up, and the potential increases proportionally. For<\/p>\nn<\/mi>=<\/mo>27<\/mn><\/mrow>n = 27<\/annotation><\/semantics><\/math>\n

drops, the potential of the large drop is:<\/p>\n

 <\/p>\nV<\/mi>large<\/mtext><\/msub>=<\/mo>V<\/mi>small<\/mtext><\/msub>\u00d7<\/mo>n<\/mi>1<\/mn>3<\/mn><\/mfrac><\/msup><\/mrow><\/semantics><\/math>\n

 <\/p>\nV<\/mi>large<\/mtext><\/msub>=<\/mo>220<\/mn>\u00d7<\/mo>2<\/mn>7<\/mn>1<\/mn>3<\/mn><\/mfrac><\/msup>=<\/mo>220<\/mn>\u00d7<\/mo>3<\/mn>=<\/mo>1980<\/mn>\u2009<\/mtext>V<\/mtext><\/mrow><\/semantics><\/math>\n

 <\/p>\n

Also Check: NCERT Solutions For Class 12 Physics Chapter 2 Electrostatics Potential<\/a><\/strong><\/em><\/p>\n

Question: What is the electric field in a region of constant potential?<\/strong><\/p>\n

(a) Uniform<\/p>\n

(b) Zero<\/p>\n

(c) Depends on the location of the charge.<\/p>\n

(d) Changes if a charge is placed inside the region.<\/p>\n

Answer: <\/strong>Zero<\/strong><\/p>\n

Explanation:<\/strong>
\nIn a region of constant potential, the electric field is zero because there is no potential gradient (<\/p>\nE<\/mi>=<\/mo>\u2212<\/mo>\u2207<\/mi>V<\/mi><\/mrow>E = -\\nabla V<\/annotation><\/semantics><\/math>\n

). This means no force acts on a charge in such a region.<\/p>\n","protected":false},"excerpt":{"rendered":"

Electrostatics is a fundamental topic in physics that deals with charges at rest and the forces, fields, and potentials associated […]<\/p>\n","protected":false},"author":53,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"Electrostatic Potential & Capacitance Class 12 MCQs","_yoast_wpseo_title":"Electrostatic Potential & Capacitance Class 12 MCQs with Answers | IL","_yoast_wpseo_metadesc":"Explore top Electrostatic Potential & Capacitance Class 12 MCQs with answers. 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