Question

Two point charges of 60.0 C and -12.0 C are separated by a distance of 20.0 cm. A 7.00 C charge is placed midway between these two charges. What is the electric force acting on this charge because of the other two charges

Answer 1

Answer:

$4.5\times 10^{14} N$

Explanation:

$q_1=60 C$

$q_2=-12 C$

$q_3=7 C$

Distance between q1 and q2,d=20 cm

$r=\frac{d}{2}=\frac{20}{2}=10 cm=10\times 10^{-2} m$

1m =100 cm

Electric force on charge q3 due to charge q1

$F_1=\frac{kq_1q_3}{r^2}=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}$ (away from q1)

Electric force on charge q3 due to charge q2

$F_2=\frac{kq_2q_3}{r^2}=\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}$(attract toward q2)

Net force=$F=F_1+F_2$

$F=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}+\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}$

$F=\frac{9\times 10^9\times 7}{(10\times 10^{-2})^2}(60+12)$

$F=4.5\times 10^{14} N$