Question

Electric field, due to an infinite line of charge, as shown in figure at a point $P$ at a disatnce $r$ from the line is $E$. If one half of the line of charge is removed from either side of point $A$, then

• A. electric field at $P$ will have magnitude $\frac{E}{2}$
• B. electric field at $P$ in $X$ direction will be $\frac{E}{2}$
• C. electric field at $P$ in $Y$ direction will be $\frac{E}{2}$
• D. none of these

Answer 1

Answer: (B), (C)

electric field at $P$ in $Y$ direction will be $\frac{E}{2}$

As we know that the elctric field due to infinite line of charge is given by
$E=\frac{2k\lambda }{r}$ ....(1)

When half of line cahrge is removed

Electric field at point $P$, in $Y$ direction:
$E_Y=\frac{k\lambda }{r}(\sin {\theta }_1+\sin {\theta }_2)$
$E_Y=\frac{k\lambda }{r}(\sin 0^{\circ }+\sin {90}^{\circ })=\frac{k\lambda }{r}$
From equation (1), $E_Y=\frac{E}{2}$

Similarly in $X$ direction:
$E_X=\frac{k\lambda }{r}(\cos {\theta }_1-\cos {\theta }_2)=\frac{k\lambda }{r}(\cos 0^{\circ }-\cos {90}^{\circ })$
$E_X=\frac{k\lambda }{r}=\frac{E}{2}$ (from equation(1))