Answer to Question #207569 in Electricity and Magnetism for DEVESH KUMAR GAUTA

Question

Charges 5.0×10−7

C,−2.5×10−7

C and 1.0×10−7

C are held fixed at the three corners A,B,C of an equilateral triangle of side 50cm respectively. Find the electric force on the charge at C due to the rest two.

Answer 1

Gives

$C_1=5\times10^{-7}C$

$C_2=-2.5\times10^{-7}C$

$C_3=1.0\times10^{-7}C$

side (a)=0.5 m

$F_{CA}=\frac{kq_3q_1}{a^2}$

$$F_{CA}=\frac{9\times10^9\times5\times10^{-7}\times1\times10^{-7}}{.5^2}=1.8\times10^{-3}N$$

$F_{CB}=\frac{kq_3q_2}{a^2}$

$$F_{CB}=-\frac{9\times10^9\times2.5\times10^{-7}\times1\times10^{-7}}{.5^2}=9\times10^{-4}N$$

$$F_{net}=\sqrt{F^2_{CA}+F^2_{CB}+2F_{CA}F_{CB}cos30°}$$

Put value

$$F_{net}=\sqrt{(1.8\times10^{-3})^2+(9\times10^{-4})^2+2\times1.8\times10^{-3}\times9\times10^{-4}\times\frac{\sqrt3}{2}}$$

$F_{net}=2.617\times10^{-3} N$