contestada

an electric field exerts a force of 3.00 E -4 N on a positive test charge of 7.20 E-4 C. The magnitude of the field at this location of the charge is

a) 0.24 N/C
b) 0.417 N/C
c) 15.6 N/C
d) 21.6 N/C
e) 2.4 N/C

Respuesta :

3.00 E-4 / 7.20 E-4 = 0.417
answer is b

Answer : Electric field, E = 0.417 N/C

Explanation :

It is given that,

Force due to charge, [tex]F=3\times 10^{-4}\ N[/tex]

Magnitude of charge, [tex]q=7.2\times 10^{-4}\ C[/tex]

We know that the electric field at a point is given by the force acting per unit charge.

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{3\times 10^{-4}\ N}{7.2\times 10^{-4}\ C}[/tex]

[tex]E=0.417\ N/C[/tex]

So, the correct option is (b) " E = 0.417 N/C "

Hence, this is the required solution.