Questions
Equipotential surfaces are shown in figure. Then the electric field strength will be
a
100Vm−1 along X -axis
b
100Vm−1 along Y -axis
c
200Vm−1 at an angle 120∘ with X -axis
d
50Vm−1 at an angle 120∘ with X -axis
detailed solution
Correct option is C
Using △V=−E→⋅r→⇒ ΔV=E.Δrsin30∘⇒ E=ΔVΔrsin30∘⇒ E=(20−10)10×10−2sin30∘⇒ E=1010×10−2sin30∘=1021/2=200V/mDirection of E→ be perpendicular to the equipotential surface and towards the decreasing value of potential i.e. at 120° (90+30)with x-axis.Talk to our academic expert!
Similar Questions
A charge of 5C experiences a force of 5000N when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of 1cm
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
Popular Books
HC Verma Part 1HC Verma Part 2RS AgarwalRD SharmaLakhmir SinghDK GoelSandeep GargTS GrewalNCERT solutions
Class 12 NCERT SolutionsClass 11 NCERT SolutionsClass 10 NCERT SolutionsClass 9 NCERT SolutionsClass 8 NCERT SolutionsClass 7 NCERT SolutionsClass 6 NCERT Solutions
