Question

Figure shows three particles with charges $q_{1}=2 Q, q_{2}=-2 Q$ and $q_{3}=-4 Q$, each a distance $d$ from the origin. Find the net electric field $\vec{E}$ at the origin.

• A. $\frac{2.56 \,Q}{4 \pi \varepsilon_{0} d^{2}}$ towards + ve x-axis
• B. $\frac{6.932 \,Q}{4 \pi \varepsilon_{0} d^{2}}$ towards + ve x-axis
• C. $\frac{6.93 Q}{4 \pi \varepsilon_{0} d^{2}}$ towards - ve x-axis
• D. Zero

Answer 1

Answer: (B)

$\frac{6.932 \,Q}{4 \pi \varepsilon_{0} d^{2}}$ towards + ve x-axis

Answer 2

The given, $q_{1}=2 Q, q_{2}=-2 Q, q_{3}=-4 Q$
The net electric field $E$ at the origin
$E=\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1}}{d^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{2}}{d^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{3}}{d^{2}}\right) \cos \theta$
$=\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 Q}{d^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 Q}{d^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q}{d^{2}}\right) \cos \theta$
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 \times 4 Q}{d^{2}} \times \frac{\sqrt{3}}{2}$
$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{8 \times 1.732 Q}{d^{2} \times 2}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{6.928 Q}{d^{2}}$
$=\frac{6.932 Q}{4 \pi \varepsilon_{0} \cdot d^{2}}$ towards $+$ ve $x$-axis.