Answer to Question #307844 in Physics for Knej

Question

Calculate the electric field experienced by a negative test charge given the following

conditions:

a. Source charge: +5.03 x 10-4 C

Distance from source charge: 2.33 x 10-5 m

b. Source charge: -4.33 x 10-12 C

Distance from source charge: 4.6 x 10-19 m

c. Source charge: +6.78 x 10-15 C

Distance from source charge: 7.9 x 10-19 m

Answer 1

The electric field is given by

$E=k\frac{q}{r^2}$

(a)

$E=9*10^9*\frac{5.03*10^{-4}}{(2.33*10^{-5})^2}=8.34*10^{15}\:\rm N/C$

(b)

$E=9*10^9*\frac{4.33*10^{-12}}{(4.6*10^{-19})^2}=1.84*10^{35}\:\rm N/C$

(c)

$E=9*10^9*\frac{6.78*10^{-15}}{(7.9*10^{-19})^2}=9.78*10^{31}\:\rm N/C$