Answer to Question #219780 in Electric Circuits for anuj

Question

Two small spheres assumed to be identical conductors are placed at 30 cm from each other on a horizontal axis. the first S1 is loaded with 12 nC and the second S2 is loaded with -18 nC. make a figure where you need representative.1) determine the electric force exerted by S1 on S2b) determine the electric field created by S1 at the level of s2c) determine the electric potential created by s1 at the level of s2d) determine the electric force exerted by S2 on S1e) determine the electric field created by s1 in the middle of [S1: S2]f) determine the electric field in the middle of [S1: S2]g) determine the electric potential in the middle of [S1: S2]h) determine the electric potential at the level of S1

Answer 1

Solution;

a) Electric field created by S₁ on S₂

$F_{s_1s_2}=\frac{k_eq_{s_1}q_{s_1}}{r^2}$ =$=\frac{8.99×10^9×12×10^{-9}×-18×10^{-9}}{0.3^2}$

=-2.157×10^-5N

b) Electric field created by S₁ at level of S₂

$E_{s_1}=\frac{k_eq_{s_1}}{r^2}=\frac{8.99×10^9×12×10^{-1}}{0.3^2}=$1198.67N/C

c) Electric potential created by s₁

$V_{s_1}=\frac{k_eq_{s_1}}{r}=\frac{8.99×10^9×12×10^{-9}}{0.3}=$ 359.6V

d) Electric force exerted by S₂ on S₁

$F_{s_2s_1}=\frac{k_eq_{s_2}q_{s_1}}{r^2}=\frac{8.99×10^9×-18×10^{-9}×12×10^{-9}}{0.3^2}=-2.157×10^{-5}N$

Negative sign means attraction force.

e) Electric field created by S₁ in the middle of S₁ and S₂

r=$\frac{0.3}{2}=0.15$ m

$E_{s_1}=\frac{8.99×10^9×12×10^{-9}}{0.15^2}=4794.67\frac NC$

f)Electric field at the middle of S₁ and S₂.

At middle ,charge is ;

$\frac{12+-18}{2}=-3nC$

E=$\frac{F_e}{q}$

$F_e=\frac{8.99×10^9×(-3×10^{-9})^2}{0.3^2}=$

3.596×10^-6N

E=$\frac{3.596×10^{-6}}{3×10^{-9}}=1189.67$ $\frac NC$

g)Electric potential in the middle.

$V=\frac{k_eq}{r}=\frac{8.99×10^9×-3×10^{-9}}{0.15}$ =179.8Volts

h) Electric potential at level of S₁ by S₂

$V=\frac{8.99×10^9×-18×10^{-1}}{0.3}$ =538.4Volts.