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

Answer:

The magnitude of force is $4.26\times 10^{- 6} N$

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:

$\vec{F} = Q\vec{E_{w}}$

$\vec{F} = -6\times 710,000 = - 4.26\times 10^{- 6} N$

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