Question

Two identical positive charges of $+Q$ are kept in a straight line at $1m$ apart. Find out the magnitude and direction of the electric field at point $A$, $0.25m$ to the right of the left-hand charge?

• A. $3/4kQ$ to the right
• B. $128/9kQ$ to the left
• C. $160/9kQ$ to the left
• D. $160/9kQ$ to the right
• E. $128/9kQ$ to the right

Answer 1

The correct option is E $128/9kQ$ to the right

Electric field at $A$ due to right hand - charge,

$E_1=k\frac{Q}{{0.75}^2}$ ........$eq1$

As it is a positive charge so direction of electric field will be away from it ,toward point $A$

Electric field at $A$ due to left hand - charge,

$E_2=k\frac{Q}{{0.25}^2}$ ........$eq2$

As it is also a positive charge so direction of electric field will be away from it ,toward point $A$

It is clear that both fields are in opposite direction at point $A$ , therefore net electric field at $A$ ,

$E=E_2-E_1$ ($E_2>E_1$ because for $E_2$ the distance is smaller from point $A$)

$E=k\frac{Q}{{0.25}^2}-k\frac{Q}{{0.75}^2}$

$E=128/9kQ$ to the right, direction of greater electric field.