Question

Charges +5 microcoulomb,10micro coulomb -10micro coulomb are placed in air at the corners a, b, c of an equilateral triangle abc, having each side equal to 5cm. determine the resultant force on the charge at a

Answer 1

the force on the charge at a is 180N

Answer 2

Answer:

The resultant force on the charge at a is 180 N.

Explanation:

Given that,

First charge $q_{1}=5\times10^{-6}\ C$

Second charge $q_{2}=10\times10^{-6}\ C$

Third charge $q_{3}=-10\times10^{-6}\ C$

The formula of force is defined as:

$F=\dfrac{kq_{1}q_{2}}{r^2}$

The force on the charge at a by the charge at b

$F_{ba}=\dfrac{9\times10^{9}\times5\times10^{-6}\times10\times10^{-6}}{(5\times10^{-2})^2}$

$F_{ba}=180\ N$

The force on the charge at a by the charge at c

$F_{ac}=\dfrac{9\times10^{9}\times5\times10^{-6}\times10\times10^{-6}}{(5\times10^{-2})^2}$

$F_{ac}=-180\ N$

Negative sign shows the attraction force.

The resultant force on the charge at a

$F_{a}=\sqrt{(F_{ba})^2+(F_{ac})^2+2\timesF_{ba}\timesF_{ac}\ cos60^{\circ}}$

$F_{a}=\sqrt{(180)^2+(180)^2+2\times180\times180\times\dfrac{-1}{2}}$

$F_{a}=180\ N$

Hence, The resultant force on the charge at a is 180 N.