Question

For the given arrangement, find the value of electric field at $C$.

• A. $2\times {10}^3\text{N/C}$
• B. $3\times {10}^3\text{N/C}$
• C. $5\times {10}^3\text{N/C}$
• D. $9\times {10}^3\text{N/C}$

Answer 1

The correct option is D $9\times {10}^3\text{N/C}$

At $C$,
Electric field due to charge at $A$,
$E_A=\frac{kq}{r^2}=\frac{9\times {10}^9\times {10}^{-8}}{{0.1}^2}=9\times {10}^3\text{N/C}$ (away from A)
Similarly, due to charge at $B$,
$E_B=9\times {10}^3\text{N/C}$ (towards B)

By principle of superposition,
$E=\sqrt{E_A^2+E_B^2+2E_A{E}_B\cos \theta }$
Here, $\theta ={120}^{\circ }$
$\Rightarrow E=\sqrt{(9\times {10}^3{)}^2+(9\times {10}^3{)}^2+2\times 9\times {10}^3\times 9\times {10}^3\times \cos {120}^{\circ }}$
$\Rightarrow E=9\times {10}^3\text{N/C}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Second method - After finding}{E}_A\text{and}{E}_B\text{at C,}\\ \text{we can take its component and add them to get resultant E.}\end{array}$

$\begin{array}{c}{E}_A\sin \theta \text{and}{E}_B\sin \theta \text{will cancel out.}\\ \text{So, net electric field,}E=E_A\cos \theta +E_B\cos \theta \end{array}$