Question

The electric potential decreases uniformly from V to -V along X-axis in a coordinate system as we moves away from a $\left( -{ x }_{ 0 },0 \right)$ to $\left( { x }_{ 0 },0 \right)$, then the electric field at the origin.
A. $Must\ be\ equal\ to\ \dfrac { V }{ { x }_{ 0 } }.$
B. $May\ be\ equal\ to\ \dfrac { V }{ { x }_{ 0 } }.$
C. $Must\ be\ greater\ than\ \dfrac { V }{ { x }_{ 0 } }.$
D. $May\ be\ less\ than\ \dfrac { V }{ { x }_{ 0 } }.$

Answer 1

Hint: To solve this problem, use the formula for electric potential which gives the relation between electric field and electric potential. Write this equation in terms of x as the motion is along the X-axis only. Then substitute the values of change in the potential and change in distance. Evaluate the obtained expression and find the electric field at the origin.

Formula used:
$ E=-\dfrac { dV }{ dr }$

Complete solution step-by-step:
Electric potential is given by,
$ E=-\dfrac { dV }{ dr }$ …(1)
Where,
$dV$ is the change in potential
$dx$ is the change in the distance
In this question, the motion is along the X-axis. So, equation. (1) can be given by,
$ { E }_{ x }=-\dfrac { dV }{ dx }$
Substituting values in above equation we get,
${ E }_{ x }=-\left( \dfrac { V-\left( -V \right) }{ \left( { -x }_{ 0 } \right) -{ x }_{ 0 } } \right) $
$\Rightarrow { E }_{ x }=-\left( \dfrac { 2V }{ -{ 2x }_{ 0 } } \right)$
$\Rightarrow { E }_{ x }=-\left( \dfrac { V }{- { x }_{ 0 } } \right)$
$\Rightarrow { E }_{ x }=\dfrac { V }{ { x }_{ 0 } }$
Thus, the electric field at the origin may be equal or greater than $ \dfrac { V }{ { x }_{ 0 } }$.

So, the correct answer is option B and C.

Note:
Students should have a clear understanding of the concept of electric field. They should know the relationship between electric field and electric potential to solve these types of questions. Students must not forget the negative sign in the formula for electric potential. The negative sign indicates that the direction of E is always in the direction of decrease of electric potential.