A positive test charge of 6.47 × 10^−5 C is placed in an electric field of 50.12 N/C intensity.

What is the strength of the force exerted onthe test charge?
Answer in units of N

A positive test charge of 7.66 × 10−5 C is

placed in an electric field of 53.01 N/C intensity.
What is the strength of the force exerted on
the test charge?
Answer in units of N.

To find the strength of the force exerted on the test charge, you can use the equation:

F = q * E

where F is the force, q is the charge, and E is the electric field intensity.

Given that the charge, q, is 6.47 × 10^−5 C, and the electric field intensity, E, is 50.12 N/C, you can substitute these values into the equation to find the force.

F = (6.47 × 10^−5 C) * (50.12 N/C)

Now, multiply the two values together to find the force:

F = 3.24 × 10^−3 N

Therefore, the strength of the force exerted on the test charge is 3.24 × 10^−3 N.