Question

The electric field intensity at a particular location surrounding a Van de Graaff generator is 4.5x103 N/C. Determine the magnitude of the force that this field would exert upon ...

Answer 1

Answer:

Explanation:

Electric field, E = 4.5 x 10^3 N/C

charge on electron, e = 1.6 x 10^-19 C

The relation between the force and the electric field is given by

F = q E

F = 1.6 x 10^-19 x 4.5 x 10^3

F = 7.2 x 10^-16 N

Answer 2

Answer:

The magnitude of the force is $7.2\times10^{-16}\ N$

Explanation:

Given that,

Electric field intensity $E= 4.5\times10^3$

Suppose, The electron when positioned at this location

We need to calculate the magnitude of the force

Using formula of electric field

$E = \dfrac{F}{q}$

$F = Eq$

Where, E = electric field

q = charge of electron

Put the value into the formula

$F=4.5\times10^3\times1.6\times10^{-19}$

$F=7.2\times10^{-16}\ N$

Hence, The magnitude of the force is $7.2\times10^{-16}\ N$