Question

Objects A and B are both positively charged. Both have a mass of 300 g , but A has twice the charge of B. When A and B are placed 60 cm apart, B experiences an electric force of 0.50 N .

a. How large is the force on A? N
b. What are the charges on qA and qB?
c. If the objects are released, what is the initial acceleration of A? (m/s^2)

Answer 1

Answer:

(a). The force on A is same as that of charge B.

(b). The charge of $q_{A}$ and $q_{B}$ is $6.324\times10^{-6}\ C$ and $3.162\times10^{-6}$.

(c). The acceleration is 1.67 m/s².

Explanation:

Given that,

Mass of both objects = 300 g

A has twice the charge of B

Distance = 60 cm

Electric force = 0.50 N

Let the charge on B is q and on A is 2q.

(a). We need to calculate the force on A

The force on charge A is also same as that of charge B due to newton's third law of motion.

(b). We need to calculate the charges on $q_{A}$ and $q_{B}$

Using formula of force

$F=\dfrac{kq_{A}q_{B}}{r^2}$

Put the value into the formula

$0.50=\dfrac{9\times10^{9}\timesQ\times2Q}{(60\times10^{-2})^2}$

$Q=\sqrt{\dfrac{0.50\times(60\times10^{-2})^2}{9\times10^{9}\times2}}$

$Q=0.000003162\ C$

$Q=3.162\times10^{-6}\ C$

The charge of $q_{A}$ is

$q_{A}= 2Q$

Put the value of Q

$q_{A}= 2\times3.162\times10^{-6}$

$q_{A}=6.324\times10^{-6}\ C$

The charge of $q_{B}$ is

$q_{B}=3.162\times10^{-6}$

(c). We need to calculate the acceleration

Using formula of acceleration

$a = \dfrac{F}{m}$

$a=\dfrac{0.50}{0.3}$

$a=1.67\ m/s^2$

Hence, (a). The force on A is same as that of charge B.

(b). The charge of $q_{A}$ and $q_{B}$ is $6.324\times10^{-6}\ C$ and $3.162\times10^{-6}$.

(c). The acceleration is 1.67 m/s².