Question

Three charges, each of magnitude 10 nC, are at separate corners of a square of edge length 3 cm. The two charges at opposite corners are positive, and the other charge is negative. Find the force exerted by these charges on a fourth charge q = +3 nC at the remaining (upper right) corner. (Assume the +x axis is directed to the right and the +y axis is directed upward.)

Answer 1

Answer:

The force exerted by three charges on the fourth is $F_{resultant}=2.74\times10^{-5}\ \rm N$

Explanation:

Given:

• The magnitude of three identical charges, $q=10\ \rm nC$
• Length of the edge of the square a=3 cm
• Magnitude of fourth charge ,Q=3 nC

According to coulombs Law the force F between any two charge particles is given by

$F=\dfrac{kQq}{r^2}$

where r is the radial distance between them.

Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition

$F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N$