Question

Part A

Consider the situation in the figure below, where two charged rods are placed a distance d on either side of an aluminum can. What does the can do?

$1343038_A.jpg$

Consider the situation in the figure below, where two charged rods are placed a distance  on either side of an aluminum can. What does the can do?

$1343038_A.jpg$

	Rolls to the right
	Rolls to the left
	Stays still

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Part B

Now, consider the situation shown in the figure below. What does the can do?

$1343038_B.jpg$

Now, consider the situation shown in the figure below. What does the can do?

$1343038_B.jpg$

	Rolls to the left
	Stays still
	Rolls to the right

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Part C

Using the setup from the first question, imagine that you briefly touch the negatively charged rod to the can. You then hold the two rods at equal distances on either side of the can. What does the can do?

Using the setup from the first question, imagine that you briefly touch the negatively charged rod to the can. You then hold the two rods at equal distances on either side of the can. What does the can do?

	Rolls toward the positively charged rod
	Rolls away from the positively charged rod
	Does not move

Answer 1

Concepts and reason

The concepts required to solve this problem are the Coulomb’s law and the charging by induction.

Initially, find the charge induced on the can due to the charged rod present on the left side of the can. Then, determine the nature of the force present between can and the rod present on the left hand side of the can. Then, find the charge induced on the can due to the charged rod present on the right side of the can. Then, determine the nature of the force present between the can and the rod on the right side. Finally, find the net force on the can. Similarly follows the above step for part B. Finally, touch the negatively charged rod to the can so, that the can gets charged and then using type of charges, determine the electrostatic force between the can and the charges.

Fundamentals

When an uncharged object is brought near to a charged object but not in contact, the uncharged object gets charged by induction from the charged object and this process of charging is known as charging by induction.

When an uncharged object is brought in contact with a charged object, the uncharged object gets charged by the transfer of electrons from the charged object and this process of the charging is known as charging by conduction.

The expression of the coulomb’s law is,

$F = k\frac{{{q_1}{q_2}}}{{{d^2}}}$

Here, F is the electrostatic force, ${q_1}$ and ${q_2}$ are the charges and d is the distance between these charges.

The electrostatic force between the like charges is repulsive in nature and the electrostatic force between the unlike charges is attractive in nature.

(A)

The charge on rod present on the left side of the can is q which is positive. Due to this positive charge, negative charge is induced on the left side of the can. So, the charge on the left side of the can is $- q$ . Therefore, the electrostatic force is attractive in nature.

The force ${F_1}$ between the charge on the rod $+ q$ present left side of the can and the charge $- q$ on the left side of the can is,

${F_1} = k\frac{{qq}}{{{d^2}}}$

The charge on rod present on the right side of the can is $- q$ which is negative. Due to this negative charge, positive charge is induced on the right side of the can. So, the charge on the right side of the can is $+ q$ . Therefore, the electrostatic force is attractive in nature.

The force ${F_2}$ between the charge on the rod $- q$ present right side of the can and the charge $+ q$ on the left side of the can is,

${F_2} = k\frac{{qq}}{{{d^2}}}$

The forces ${F_1}$ and ${F_2}$ are equal in magnitude but opposite in direction. So, the net force on the can is zero.

Thus, the can neither rolls to the left nor to the left. It remains still, does not move.

(B)

The charge on rod present on the left side of the can is q which is positive. Due to this positive charge, negative charge is induced on the left side of the can. So, the charge on the left side of the can is $- q$ . Therefore, the electrostatic force is attractive in nature between them.

The force ${F_1}$ between the charge on the rod $+ q$ present left side of the can and the charge $- q$ on the left side of the can is,

${F_1} = k\frac{{qq}}{{{d^2}}}$

The charge on rod present on the right side of the can is $q$ which is positive. Due to this positive charge, negative charge is induced on the right side of the can. So, the charge on the right side of the can is $- q$ . Therefore, the electrostatic force is attractive in nature.

The force ${F_2}$ between the charge on the rod $+ q$ present right side of the can and the charge $- q$ on the left side of the can is,

${F_2} = k\frac{{qq}}{{{d^2}}}$

The forces ${F_1}$ and ${F_2}$ are equal in magnitude but opposite in direction. So, the net force on the can is zero.

Thus, the can neither rolls to the left nor to the left. It remains still, does not move.

(C)

Refer the figure given in part A of the question.

A positive charged rod is present on the left side of the can and a negative charged rod is present on the right side of the can.

When a negatively charged rod is brought in contact with the can, the can gets negatively charged due to the contact with the negatively charged rod due to charging by conduction.

Now, the can is negatively charged. The can will be attracted to the positively charged rod present on the left side of the can and the can will be repulsive in nature to the negatively charged rod present on the right side of the can.

Thus, the can neither stays still nor move towards the negatively charged rod. The can will rolls toward the positively charged rod due to attractive force present between them.

Ans: Part A

The can stays still.