Question

The electric potential due to a point charge decreases uniformly from $120\text{V}$ to $80\text{V}$ as one moves on the $x-$ axis from $x=-1\text{cm}$ to $x=+1\text{cm}$. The electric field at the origin.

• A. must be equal to $20\text{V/cm}$
• B. must be equal to $20\text{V/m}$
• C. may be greater than $20\text{V/cm}$
• D. may be less than $20\text{V/cm}$

Answer 1

The correct option is C may be greater than $20\text{V/cm}$

Given,

Electric potential at $x_1=-1\text{cm}$ is $V_1=120\text{V}$

Electric potential at $x_2=+1\text{cm}$ is $V_2=80\text{V}$

Since the potential is uniformly decreasing, we can write that,

$E_x=-\left(\frac{V_2-V_1}{x_2-x_1}\right)$

$\Rightarrow E_x=-\left(\frac{80-120}{1-(-1)}\right)=+20\text{V/cm}$

Since, $E_x$ is $x-$ component of electric field vector. The effect of $y-$ component and $z-$ component will also be at origin.

Thus, the electric field at origin may be greater than $+20\text{V/cm}$

Hence, option (c) is the correct answer.