Question:

Two point charges $ +8q$ and $-2q$ are placed as shown in the figure.
The point where the electric field is zero is

Updated On: Aug 1, 2022
  • $x = l$
  • $x = 2l$
  • $x = 3l$
  • $x = 4l$
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The Correct Option is B

Solution and Explanation

Let at a point $P$ at a distance $d$ from $-2q$ charge at which the electric field is zero.
$\therefore E_1 = E_2$ $\frac{1}{4\pi\varepsilon_{0}} \frac{8q}{\left(l+d\right)^{2}} = \frac{1}{4\pi\varepsilon_{0}} \frac{2q}{d^{2}}$ $ 4d^{2} = \left(l+d\right)^{2} $ $ 2d = l+d $ $d= l$ Hence, at $x = 2l$, the electric field is zero due to given charge configuration.
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Notes on Electric Field

Concepts Used:

Electric Field

Electric Field is the electric force experienced by a unit charge. 

The electric force is calculated using the coulomb's law, whose formula is:

\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)

While substituting q2 as 1, electric field becomes:

 \(E=k\dfrac{|q_{1}|}{r^{2}}\)

SI unit of Electric Field is V/m (Volt per meter).