Question

If Electric field in a region is given by $E=x^2\hat{i}+y\hat{j}+\hat{k}$. What is the potential of the point (1,2,3) given the potential of the origin is 10V?

• A. $\frac{16}{3}V$
• B. $-\frac{16}{3}V$
• C. $\frac{14}{3}V$
• D. $-\frac{14}{3}V$

Answer 1

The correct option is C

$\frac{14}{3}V$

Since we know $E=-\frac{dV}{dr},$
To find V, we use the inverse operation of differentiation, which is integration.
$V=-{\int }_{r_1}^{r_2}Edr$
So in a Cartesian coordinate system, the above equations become
$E=-(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k})$
Where, $\frac{\partial }{\partial x}$represents the partial derivative w.r.t x alone keeping variables y and z constant.
So for the above question,
$V=-({\int }_0^1{x}^2+{\int }_0^{2}y+{\int }_0^{3}1)$
$V=-(\frac{x^3}{3}{|}_0{}^1+\frac{y^2}{2}|0{}^2+z|0{}^3)$
$V=-(\frac{1}{3}+2+3)$
$V=-\frac{16}{3}V$

Remember that this is only the potential difference between points (1,2,3) and (0,0,0).
Since the potential at (0,0,0) is 10 V,
The answer we are looking for is $V=10-\frac{16}{3}=\frac{14}{3}.$