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Physics Question on Electric Field

The electric intensities at a point due to two point charges in the x−yx - y plane are 3i^−2j^3 \hat{i} -2 \hat{j} and −2i^+4j^-2\hat{i} +4\hat{j} The magnitude of the resultant intensity at that point is

A

2.08 V m−12.08 \,V\,m^{-1}

B

2.24 V m−12.24 \,V\,m^{-1}

C

1.8 V m−11.8 \,V\,m^{-1}

D

3.5 V m−13.5 \,V\,m^{-1}

Answer

2.24 V m−12.24 \,V\,m^{-1}

Explanation

Solution

Given : E⃗1=3i^−2j^,E⃗2=−2i^+4j^\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 E⃗=E⃗1+E⃗2=(3i^−2j^)+(−2i^+4j^)=i^+2j^\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} ∣E⃗∣=(1)2+(2)2\left|\vec{E}\right|=\sqrt{\left(1\right)^{2}+\left(2\right)^{2}} =1+4=5=2.24 V m−1=\sqrt{1+4}=\sqrt{5}=2.24\,V\,m^{-1}