Question

The potential at a point x due to some charge is given by equation V(x) = $\frac{20}{(x^{2} - 4)}$ volts. Then electric field at x =4 is given by –

• A. $\frac{5}{3}$ volt/m and in the –ve x direction
• B. $\frac{5}{3}$volt/ m and in the +ve x direction
• C. $\frac{10}{9}$ volt/m and in the –ve x direction
• D. $\frac{10}{9}$ volt/m and in the +ve x direction

Answer 1

Answer: (D)

$\frac{10}{9}$ volt/m and in the +ve x direction

Answer 2

E_x = –= – $\frac { \delta } { \delta x }$ $\left( \frac { 20 } { x ^ { 2 } - 4 } \right)$ = $\frac { 20 \times 2 \mathrm { x } } { \left( \mathrm { x } ^ { 2 } - 4 \right) ^ { 2 } }$

E_x = $\frac { 40 x } { \left( x ^ { 2 } - 4 \right) ^ { 2 } }$ = $\frac { 40 \times 4 } { 144 }$ =$\frac { 10 } { 9 }$ in + x direction.