Question

A small metal ball with a mass of 10 grams and a charge of 5.0 mC is located in a region of space with an electric field of 12 N/C directed to the east. The ball is released from rest at a distance of 30 cm above the ground. What is the speed of the ball just before it hits the ground?

Answer 1

I am not sure about "a" whether we should take g or the one I got please do tell if it's wrong

Answer 2

The speed of the ball just before it hits the ground is 1.9 m/s.

Given,

Mass of the ball=10 g

Charge on the ball=5 mC

Electric field around the ball=12 N/C

Height above the ground from where it was released=30 cm.

To find,

the speed of the ball just before it hits the ground.

Solution:

• The force on a charge q in an electric field E is given as per the following expression:
• $F=qE$.
• The force on a body of mass m moving with an acceleration a is given as:
• $F=ma$.

The force on a charge q of mass m moving with an acceleration a in an electric field E will be:

$qE=ma\\a=\frac{qE}{m} .$

The acceleration of the ball will be:

$a=\frac{qE}{m} \\a=\frac{5 *10^{-3} *12}{\frac{10}{1000} } \\a=\frac{5 *10^{-3} *12}{0.01}\\a=5 *10^{-3} *12*100\\a=5 *10^{-1} *12\\a=6 m/s^{2} .$

The initial speed, u of the ball will be zero(as it was released).

The speed of the ball when it hits the ground will become zero.

Let the speed of the ball just before it hits the ground be v.

According to equation of motion,

$v^{2} =u^{2} +2as\\v^{2}=0+2(6)(\frac{30}{100} )\\v^{2}=3.6 \\v=\sqrt{3.6} \\v=1.89 m/s\\v=1.9 m/s.$

Thus, the speed of the ball will be 1.9 m/s.

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