Question

A 3.00 × 10^−9-coulomb test charge is placed near

a negatively charged metal sphere. The sphere

exerts an electrostatic force of magnitude

6.00 × 10^−5 newton on the test charge. What is

the magnitude and direction of the electric field

strength at this location?

(1) 2.00 × 10^4 N/C directed away from the

sphere

(2) 2.00 × 10^4 N/C directed toward the sphere

(3) 5.00 × 10^−5 N/C directed away from the

sphere

(4) 5.00 × 10^−5 N/C directed toward the sphere

Answer 1

(2) 2.0×10⁴ N/C directed towards the sphere

Step-by-step explanation:

Electric Field: This can be defined as the force per unit charge. The S.I unit of Electric Field is N/C.

The expression for electric Field is given as,

E = F/q...................... Equation 1

Where E = Electric Field, F = Force, q = charge.

Given: F = 6.0×10⁻⁵ N, q = 3×10⁻⁹ C

Substitute into equation equation 1

E = 6.0×10⁻⁵/(3×10⁻⁹)

E = 2.0×10⁴ N/C directed towards the sphere

Hence the right option is (2) 2.0×10⁴ N/C directed towards the sphere

Answer 2

B

Step-by-step explanation:

Given:-

- The charge of the test particle q = 3.0 * 10^-9 C

- The force exerted by the metal sphere F = 6.0 * 10^-5 N

Find:-

The magnitude and direction of the electric field strength at this location?

Solution:-

- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:

F = E*q

- Using the data given we can determine E:

E = F / q

E = (6.0 * 10^-5) / (3.0 * 10^-9)

E = 20,000 N/C

- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.