Question

Along the positive direction of x-axis there exists a constant electric field $E_0$. If the electric potential at $x=0$ is zero, then its value at a point $P$, positioned at ${}^{\prime }{x}^{\prime }\text{m}$ wrt origin will be

• A. $V\left(x\right)=+x{E}_0$
• B. $V\left(x\right)=-x{E}_0$
• C. $V\left(x\right)=+x^2{E}_0$
• D. $V\left(x\right)=-x^2{E}_0$

Answer 1

The correct option is B $V\left(x\right)=-x{E}_0$

As we know that the relation between electric field and potential difference is given by,

$E=-\frac{dV}{dr}$

Now potential between $x=0\text{and}x=+x$

${\int }_0^{V}dV=-{\int }_0^x{E}_{0}dr$

$V=-E_0{\int }_0^{x}dr=-E_0[r{]}_0^x$

$V=-E_{0}x$