Question:

Two point charges $+ 8\, q$ and $- 2\, q$ are located at $x = 0$ and $x = L$ respectively. The location of a point on the x-axis at which the net electric field due to these two point charges is zero is :

Updated On: Aug 1, 2022
  • $2\,L$
  • $\frac{L}{4}$
  • $8L$
  • $4\,L$
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The Correct Option is A

Solution and Explanation

Suppose that a point $B$ , where net electric field is zero due to charges $8q$ and-$2$ $\overrightarrow{E}_{BO}$$=\frac{-1}{4\,\pi\,\varepsilon_{0}}. \frac{8q}{a^{2}}\hat{i}$ $\overrightarrow{E}_{BA}$$=\frac{1}{4\,\pi\,\varepsilon_{0}}. \frac{+2q}{\left(a+L\right)^{2}}\hat{i}$ According to condition $\overrightarrow{E}_{BO}$ +$\overrightarrow{E}_{BA}=0$ $\therefore \frac{-1}{4\,\pi\,\varepsilon_{0}}. \frac{8q}{a^{2}}=\frac{1}{4\,\pi\varepsilon_{0}} \frac{2q}{\left(a+L\right)^{2}}$ $\Rightarrow \frac{2}{a}=\frac{1}{a+L}$ $\Rightarrow 2a+ 2L = a$ $\therefore 2L=-a$ Thus, at distance $2L$ from oxigin, net electric field will be zero.
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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).