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

Answer 2

The magnitude of the electric field and the direction of the electric filed acting on the object is,

• (1) The magnitude of the electric field is $2.05\times10^4\rm N/C$
• (2) The direction of the electric field is downward.
• (3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude is downward and acceleration is 9.8 m/s² respectively.

What is electric field?

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

The mass of the object is 0.0044 kg, and the charge on the object is -2.1   μC.

• (1) The magnitude of the electric field-

As the electric field is the ratio of force (due to gravity in given case) to the charge given on it. Thus the magnitude of the electric field of 0.0044 kg object is,

$E=\dfrac{0.0044\times9.8}{2.1\times10^{-6}}\\E=2.05\times10^{4}\rm N/C$

• (2) The direction of the electric field-

The electric force acting on the body is in the upward direction Thus, to balance this, the electric field acts on the body is in the downward direction.

(3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude of its acceleration-

When the electric charge on the object is doubled while its mass remains the same, then the strength of the electric field is halved. Therefore,

$E=\dfrac{2.05\times10^4}{2}\\E=1.025\times10^4\rm N/C$

The electric force acting on the body is in the upward direction Thus, to balance this, the electric field acts on the body is in the downward direction.

Hence, the magnitude of the electric field and the direction of the electric filed acting on the object is,

• (1) The magnitude of the electric field is $2.05\times10^4\rm N/C$
• (2) The direction of the electric field is downward.
• (3) The electric charge on the object is doubled while its mass remains the same, then the direction and magnitude is downward and acceleration is 9.8 m/s² respectively.

Learn more about electric field here;

https://brainly.com/question/14372859