Question

An object with a charge of −2.1 μC and a mass of 0.0044 kg experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. (a) Find the direction and magnitude of the electric field. (b) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its
(1) Find the magnitude of the electric field.
(2) Find the direction of the electric field.
(3) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.
(4) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.

Answer 1

Answer:

(1) 2.05 x 10^4 N/C

(2) Downward

(3) upward, 9.8 m/s^2

(4) upward, 9.8 m/s^2

Explanation:

q = - 2.1 micro coulomb, m = 0.0044 kg, g = 9.8 m/s^2

(1) The electric force is given by F = - q x E

The magnitude of electric force is balanced by the weight of the charged particle

q x E = m x g

E = mg / q

$E = \frac{0.0044 \times 9.8}{2.1 \times 10^{-6}}$

E = 2.05 x 10^4 N/C

(2) As the electric force is acting upward and the weight is downward so the elecric field is in downward direction.

(3) The charge is doubled,

then the electric field becomes half.

E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C

The direction is same that is in downward direction.

Acceleration = Force / mass

a = mg / m = 9.8 m/s^2 upward

(4) The charge is doubled,

then the electric field becomes half.

E = 2.05 x 10^4 / 2 = 1.025 x 10^4 N/C

The direction is same that is in downward direction.

Acceleration = Force / mass

a = mg / m = 9.8 m/s^2 upward