Question:
Two point charges $ + 10^{-7} C $ and $ - 10^{-7} C $ are placed at $ A $ and $ B, 20 \,cm $ apart as shown in the figure. Calculate the electric field at $ C, \,20\, cm $ apart form both $ A $ and $ B $ .
Two point charges $ + 10^{-7} C $ and $ - 10^{-7} C $ are placed at $ A $ and $ B, 20 \,cm $ apart as shown in the figure. Calculate the electric field at $ C, \,20\, cm $ apart form both $ A $ and $ B $ .
Updated On: Jun 14, 2022
- $ 1.5 \times 10^{-5} N/C $
- $ 2.2 \times 10^{4} N/C $
- $ 3.5 \times 10^{6} N/C $
- $ 3.0 \times 10^{5} N/C $
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The Correct Option is B
Solution and Explanation
The magnitude of each electric field vector at point $C$ due to charge $q_1$ and $q_2$
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}$
$E = \frac{9 \times 10^9 \times 10^{-7}}{(20 \times 10^{-2})^2}$
$ = 2.2 \times 10^4 \,N/C$
$E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}$
$E = \frac{9 \times 10^9 \times 10^{-7}}{(20 \times 10^{-2})^2}$
$ = 2.2 \times 10^4 \,N/C$
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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).