The intensity of the electric field because of a spherical shell with a thin uniformly charged surface: The unit positive electric charge is subjected to force F. The electric field is described by q at a location. E=F/q is the formula for calculating the electric field. The force exerted on a negative charge is in the opposite direction as the force exerted on a positive charge. The electric field is a two-dimensional vector quantity with both magnitude and direction. Field lines, also known as force lines, are used to depict the electric field.
Let’s take a closer look at the electric field intensity produced by an evenly charged spherical shell in this article. The intensity of the electric field Because of a spherical shell with a thin uniformly charged surface
We have three examples in this derivation.
- Outside the shell, there is an electric field.
- Inside the shell, there is an electric field.
- At the surface, there is an electric field.
The electric field can be seen as a collection of lines with the same direction at each location as the field, an idea presented by Michael Faraday, whose name ‘lines of force’ is still used occasionally. The force of the field in this illustration is linked to the number of lines, which is an important property. The field lines are indeed the routes that a point positive charge might adopt if it were forced to move inside the field, analogous to the trajectories that masses would make if they were forced to relocate inside a gravitational field. Field lines caused by stationary charges have numerous significant qualities, including that they always start with positive charges and end with negative charges, that they enter all suitable conductors at right angles, and that they never intersect or close in on themselves. The field lines symbolize an idea; the field truly pervades all of the space in between the lines. Depending on the level of precision required to depict the field, a little more and fewer lines may be created. Electrostatics is the study of electric fields formed by stationary charges.
A brief outline:
The following three examples are examined when determining the electric field due to an evenly charged thin spherical shell:
- Case 1: r > R at a point outside the spherical shell.
- Case 2: At a point on a spherical shell’s surface where r = R.
- Case 3: Inside the spherical shell, at a position where r < R.
Outside the Spherical Shell, there is an electric field:
Electric Field is the force experienced by a unit positive charge or a test charge when it is kept near a charge. It’s also known as the area that attracts or repels a charge. E stands for the electric field, which is a vector quantity. The electric field is measured in N/C, which is the standard unit. We take point P outside the spherical shell at a range of r from the center of the spherical shell to evaluate the electric field outside the shell. We are using a Gaussian spherical shape with radius r and center O for symmetry. As all points are similarly spaced, the Gaussian surface will pass through P and encounter a constant electric field of E all around.
Inside the Spherical Shell, there is an electric field:
Electric Field is the force felt by a unit positive charge or a test charge when it is kept near a charge. It’s also known as the area that draws or repels a charge. E stands for the electric field, which is a vector quantity. The electric field is measured in N/C units.
Take a point P within the spherical shell to measure the electric field inside the shell. We can use symmetry to create a spherical Gaussian surface that passes through P, is centered at O, and has a radius of r.
Electric Field at the Surface of The Shell:
Gauss’s law quickly produces a well-known discontinuity in the electric field for a charged conducting sphere. In literature, including some textbooks, different results are found at a point on the sphere’s surface, where a ‘half-factor’ multiplies the maximum strength Qk/R2, achieved just outside the sphere. Unconvinced by the existing arguments for the existence of this ‘half-factor,’ all of which are based on Gauss’s law.
Outside The Shell, There Is an Electric Field:
Consider the case of a point P outside of the spherical shell. OP = r in this case. The radius of the Gaussian surface as just a sphere is presumed to be r, as indicated in the diagram below. At every point on a Gaussian surface pointed outwards, the electric field intensity, E, is stated to be the same.
In the above-mentioned Gaussian surface, since the electric field is directed outwards (as in-unit vector ^n
Θ = 00 ∮s E. ds = ∮s E. n .ds = q/ϵ0
E∮d S = q/ϵ0 E = 4πr 2 = q/ϵ0
E = q/4πr 2 ϵ0
We can see that the electric intensity at any place outside of the spherical shell seems to be concentrated around the shell’s midpoint.
Inside The Shell, There Is an Electric Field:
Consider the position of point P within the shell. The Gaussian surface has radius r, as indicated in the diagram above. There is no charge on the Gaussian surface.
As a result,
E = 0
As a result, the field within the spherical shell must always be zero.
Electric Field at The Surface of The Shell:
Consider, r = R.
Ponder the charge density on the shell to be σ cm-2
q=4πR 2 * σq
= 4πR 2.σ / σ4πR 2.ϵ0
Significance of uniformly charged thin spherical shell:
The electric field lines NEET ponderings are expected to disclose and give replies to the most often posed inquiries on the test. With the assistance of notes from proficient researchers in the field, which is given on the Infinity Learn online stage, these can be clarified in basic terms. Assuming understudies take care of an exhaustive appreciation of the subjects all through the educational program, numerous decision questions are obvious to rehearse. How many Important NEET questions help understudies to plan for a considerable length of time questions, that are predominant in the vital test.
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FAQs (Frequently asked questions)
Question 1. Inside the spherical shell, what really is the electric field?
Question 2. What would be the electric field at the spherical shell’s surface?
Answer: q = 4πR 2.σ/ 4πR 2.ϵ0 = σ/ϵ0
Question 3. What would be the electric field outside the spherical shell’s surface?
Answer: E = q/4πr 2ϵ0
Question 4. How do you calculate the electric field?
Question 5. Is the electric field a scalar or a vector quantity?
Answer: A vector quantity is an electric field.
Question 6. What are all the 3 kinds of symmetry that must be taken into account when calculating the electric field?
Answer: The three kinds of symmetry to consider while determining the electric field are as follows:
- Symmetry on a plane
- Symmetry in a cylindrical shape
- Symmetry on a sphere
Question 7. Is the Gauss law applicable to any closed or open surface?
Answer: The Gauss law only applies to closed surfaces; it does not apply to open surfaces.