Question

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is 3.5 µC, and the lengths of the sides of the triangle are 2.9 cm. Calculate the magnitude of the net force that each charge experiences.

Answer 1

Answer:

The magnitude of the force on positive charges will be $\bf{227.06~N}$ and the magnitude of the force on the negative charge is $\bf{302.7~N}$.

Explanation:

Given:

The value of the charges, $q = 3.5~\mu C$.

The length of each side of the triangle, $l = 2.9~cm$.

Consider a equilateral triangle $\bigtriangleup ABC$, as shown in the figure. Let two point charges of magnitude $q$ are situated at points $A$ and $B$ and another point charge $-q$ is situated at point $C$.

The value of the force on the charge at point $A$ due to charge at point $C$ is given by

$F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA$

The value of the force on the charge at point $A$ due to charge at point $B$ is given by

$F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA$

The net resultant force on the charge at point $A$ is given by

$~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)$

The value of the force on the charge at point $B$ due to charge at point $C$ is given by

$F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB$

The value of the force on the charge at point $B$ due to charge at point $A$ is given by

$F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB$

The net resultant force on the charge at point $B$ is given by

$~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)$

The value of the force on the charge at point $C$ due to charge at point $A$ is given by

$F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC$

The value of the force on the charge at point $C$ due to charge at point $B$ is given by

$F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC$

The net resultant force on the charge at point $C$ is given by

$F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)$

Substitute $3.5~\mu C$ for  $q$ , $0.029~m$ for  $l$  and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (1), we have

$F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for  $q$ , $0.029~m$ for  $l$  and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (2), we have

$F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N$

Substitute $3.5~\mu C$ for  $q$ , $0.029~m$ for  $l$  and $9 \times 10^{9}~Nm^{2}C^{-2}$ for $k$ in equation (3), we have

$F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N$