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

0.0175 N

8750000 N/C

Step-by-step explanation:

F = Force = 0.07 N

E = Electric field

Electric force is given by

`F=Eq\\\Rightarrow E=(F)/(q)\\\Rightarrow E=(0.07)/(8* 10^(-9))\\\Rightarrow E=8750000\ N/C`

The electric field is 8750000 N/C

For 2 nC

`F=Eq\\\Rightarrow F=8750000* 2* 10^(-9)\\\Rightarrow F=0.0175\ N`

The force is 0.0175 N