Question

In a certain region of space with volume 0.2\[{m^3}\], the electric potential is found to be 5V throughout. The magnitude of the electric field in this region is :
A. 0.5 N/C
B. 1 N/C
C. 5 N/C
D. Zero

Answer 1

Hint: Electric field is defined as the region of influence around a charge where force is experienced by other charged particles in that region. It is a vector quantity. The electric field is equal to the negative gradient of electric potential. Electric potential is defined as the electric potential energy per unit charge which is the amount of work needed to bring a unit positive charge from infinity to a particular point. It is a scalar quantity.

Formula used:
\[E = - \dfrac{{\partial V}}{{\partial r}}\]
where E = Electric field and V = Electric Potential

Complete step by step solution:
Given: V = 5 V constant throughout the region and volume of space = 0.2 \[{m^3}\]
We know,
\[E = - \dfrac{{\partial V}}{{\partial r}}\]
However, it is given that the potential is constant throughout the given volume of space.
Therefore, \[E = - \dfrac{{\partial V}}{{\partial r}}\]= 0

When a charge is moved from one point to another, it is called a potential difference. The electric field is present only in case of a potential difference. In the case of regions with no potential difference, the net work done in moving a charge as per the definition of potential becomes zero. Hence there is no force exerted for a given displacement. Thus, there is no electric field or the electric field is zero.

Similarly, if no displacement of a charged particle takes place in the region with varying potential, then also the electric field will be zero because no work is done as there is no displacement.

Hence (D) is the correct option.

Note: Alternatively, if the potential is constant everywhere, then the surface becomes equipotential. The electric field is zero for an equipotential surface.