Question:
The electric intensities at a point due to two point charges in the $ x - y $ plane are $3 \hat{i} -2 \hat{j} $ and $ -2\hat{i} +4\hat{j} $ The magnitude of the resultant intensity at that point is
The electric intensities at a point due to two point charges in the $ x - y $ plane are $3 \hat{i} -2 \hat{j} $ and $ -2\hat{i} +4\hat{j} $ The magnitude of the resultant intensity at that point is
Updated On: Jul 29, 2022
- $ 2.08 \,V\,m^{-1} $
- $ 2.24 \,V\,m^{-1} $
- $ 1.8 \,V\,m^{-1} $
- $ 3.5 \,V\,m^{-1} $
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The Correct Option is B
Solution and Explanation
Given : $\vec{E}_{1}=3 \hat{i}-2 \hat{j}, \vec{E}_{2}=-2\hat{i}+4\hat{j}$
Resultant electric intensity at given point is
$\vec{E}=\vec{E}_{1}+\vec{E}_{2}=\left(3\hat{i}-2 \hat{j}\right)+\left(-2 \hat{i}+4 \hat{j}\right)=\hat{i}+2\hat{j}$
$\left|\vec{E}\right|=\sqrt{\left(1\right)^{2}+\left(2\right)^{2}}$
$=\sqrt{1+4}=\sqrt{5}=2.24\,V\,m^{-1}$
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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).