Question

When a test charge of 8.00 nC is placed at a certain point the force that acts on it has a magnitude of 0.0700 M and is directed northeast 1 If the test charge were 2.00 nC instead,

(a) What would be the magnitude of the force that would act on it ?
(b) What is the electric field at the point in the question?Express your answer to three significant figures

Answer 1

Answer:

0.0175 N

8750000 N/C

Explanation:

F = Force = 0.07 N

E = Electric field

Electric force is given by

$F=Eq\\\Rightarrow E=\dfrac{F}{q}\\\Rightarrow E=\dfrac{0.07}{8\times 10^{-9}}\\\Rightarrow E=8750000\ N/C$

The electric field is 8750000 N/C

For 2 nC

$F=Eq\\\Rightarrow F=8750000\times 2\times 10^{-9}\\\Rightarrow F=0.0175\ N$

The force is 0.0175 N