Question

The electric-field strength vector is given by2.5 xi 1.5y). The electric potential at point(2m , 2m, 1m) is (Considering electric potential at origin to be zero)

Answer 1

Step 1: Given that:

The electric field in vector form is given as,

$\overrightarrow{E}=2.5x\hat{i}+1.5y\hat{j}$

The coordinate of the given point = $(2m,2m,1m)$

Step 2: Calculation of the electric potential:

We have, the electric potential is given as,

$-V=\int \overrightarrow{E}.\overrightarrow{dr}$

Now,

Let the position vector of the place at which the potential is to be determined;

$\overrightarrow{dr}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

Now, using the values, we have;

$-V=\int (2.5x\hat{i}+1.5y\hat{j}).(dx\hat{i}+dy\hat{i}+dz\hat{k})$

$-V=\int (2.5xdx+1.5ydy)$

Now, integrating the above integrand in the limits, that is x component of position varies from 0 to 2m and y component of position varies from 0 to 2m, we have;

$-V={\int }_{(0,0)}^{(2m,2m)}(2.5xdx+1.5ydy)$

$-V={\int }_0^{2}2.5xdx+{\int }_0^{2}1.5ydy$

$-V=2.5{\left[\frac{x^2}{2}\right]}_0^2+1.5{\left[\frac{y^2}{2}\right]}_0^2$

$-V=2.5[\frac{2^2}{2}-0]+1.5[\frac{2^2}{2}-0]$

$-V=2.5\times 2+1.5\times 2$

$-V=5+3$

$V=-8V$

Thus,

The electric potential at the given point is $-8V$ .