First slide

Electrostatic force

Question

Five point charges, each of value +q, are placed on five vertices of a regular hexagon of side L. The magnitude of the force on a point charge of value -q coulomb placed at the center of the hexagon is

Moderate

A

$\frac{1}{π ε_{0}} {(\frac{q}{L})}^{2}$
B

$\frac{2}{π ε_{0}} {(\frac{q}{L})}^{2}$
C

$\frac{1}{2 π ε_{0}} {(\frac{q}{L})}^{2}$
D

$\frac{1}{4 π ε_{0}} {(\frac{q}{L})}^{2}$

Solution

If there had been a sixth charge +q at the remaining vertex of hexagon, force due to all the six charges on -q at O will be zero. Now if $f$ is the force due to the sixth charge and F due to the remaining five charges, then
$\begin{array}{c} \vec{F} + \vec{f} = 0, i.e., \vec{F} = - \vec{f} \\ \Rightarrow | \vec{F} | = | \vec{f} | = \frac{1}{4 π ε_{0}} \frac{q \times q}{L^{2}} = \frac{1}{4 π ε_{0}} {(\frac{q}{L})}^{2} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App