Question

An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places

Answer 1

The magnitude of force is
`4.26* 10^(- 6) N`

Solution:

As per the question:

The strength of Electric field due west at a certain point,
`\vec{E_(w)} = 710,000 N/C`

Charge, Q = - 6 C

Now, the force acting on the charge Q in the electric field is given by:

`\vec{F} = Q\vec{E_(w)}`

`\vec{F} = -6* 710,000 = - 4.26* 10^(- 6) N`

Here, the negative sign indicates that the force acting is opposite in direction.