Question

What is the direction and magnitude of the electric field at point $x_0$ halfway in between the two negative charges?

• A. $\frac{-Q_0}{3\pi {ϵ}_0{d}^2}\hat{i}$
• B. $\frac{Q_0}{4\pi {ϵ}_0{d}^2}\hat{i}$
• C. $\frac{Q_0}{3\pi {ϵ}_0{d}^2}\hat{i}$
• D. $\frac{-Q_0}{4\pi {ϵ}_0{d}^2}\hat{i}$

Answer 1

The correct option is A $\frac{-Q_0}{3\pi {ϵ}_0{d}^2}\hat{i}$

Let the distance between the point $x_o$ and the positive $Q_0$ charge

Using Pythagoras theorem we get

$d^2=\frac{d^2}{4}+K^2$

$d^2-\frac{d^2}{4}=K^2$$\Rightarrow \frac{3d^2}{4}$

Now the electric field die to $-{\theta }_o$ and $-{\theta }_o$ cancel out each other and the resultant electric field is due to the $+{\theta }_o$

$\therefore E=\frac{K{\theta }_o}{K^2}=\frac{1\times (+{\theta }_o)}{4\pi {ϵ}_o\times 3d^2/4}$

$=\frac{{\theta }_o}{3\pi {ϵ}_o{d}^2}$

($-j$ signifies that it is directed in $-k$ direction).