Question

An electric field →E=→i Ax exists in space, where A = 10 V m−2. Take the potential at (10 m, 20 m) to be zero. Find the potential at the origin.

Answer 1

At the origin, the potential is 500 V.

Explanation:

Step 1:

Given:

Electric field intensity,

E→=iAx

=10 x $i^{potential}$

where A = 10 V $m^{-2}$.

Step 2:

Take the potential at (10 m, 20 m) to be zero.

E = 10 x $i^{potential}$

E = 10 x $i^0$

$i^0$ = 1

So

E = 10x

E = -dV / dx

-dV / dx = 10x

dV = -E→. dx→ = -10x dx

Step 3:

On integrating both sides,

we get

$\begin{equation}\begin{array}{c}{\int d V=-E \rightarrow, \int d x \rightarrow=\int-10 x d x} \\{V=\frac{10 \times x^{2}}{2}} \\{V=\frac{10 \times x^{2}}{2}}\end{array}$

                   V = 5 × $x^2$

Step 4:

Take the potential at (10 m, 20 m)

X = 10 m

$\begin{equation}\begin{aligned}&V=5 \times 10^{2}\\&V=5 \times 100\\&V=500 \mathrm{V}\end{aligned}$

At the origin, the potential is 500 V.