High School Physics

# Numericals on Coulomb’s Law class 12

Here you will find a set of Numericals on Coulomb’s Law for class 12. This consists of easy-to-hard Numerical questions for Physics class 12.

Formulas Used

|F| = k q1q2/r2

Numericals on Coulomb’s Law for class 12

Question 1] A small metal sphere with a negative charge of 2.10 x 10-6 C is brought near an identical sphere with a positive charge of 1.50 x 10-6 C so that the distance between the centers of the two spheres is 3.30 cm (see Figure). Calculate the magnitude and type (attraction or repulsion) of the force of one charge acting on another.

Solution:

given:
q1 = – 2.10 x 10-6 C
q2 = + 1.50 x 10-6 C
r =3.3 x 10-2 m

To find out: the magnitude and type of the electrostatic force acting on the two charges

The magnitude of the electrostatic force is |Fe| = k q1q2/r2
=> |Fe|= 9 x 109 x (2.10 x 10-6 )(1.50 x 10-6)/(3.3 x 10-2)2 = 26 N

The charges have opposite signs, so the force is attractive

Note:
The magnitude calculation does not use the positive and negative signs for the charges.
However, we can use these signs to determine whether the electrostatic force is attractive or repulsive.

Question 2] The two identical spheres of question 1 above, one with a negative charge of 2.10 x 10-6 C and the other with a positive charge of 1.50 x 10-6 C, are momentarily brought together and then returned to their original separation distance of 3.30 cm. Determine the electrostatic force now exerted by one charge on the other.

Solution:

Initial condition:
q1 = – 2.10 x 10-6 C
q2 = + 1.50 x 10-6 C
r =3.3 x 10-2 m

When a sphere with a negative charge of – 2.10 x 10-6 C momentarily touches a sphere with a positive charge of + 1.50 x 10-6 C, then -1.50 x 10-6 C of charge from the first sphere neutralizes the +1.50 x 10-6 C of charge on the second sphere. The remaining charge of -0.60 x 10-6 C from the first sphere then divides equally between the two identical spheres. Each sphere now has a charge of – 0.3 x 10-6 C.

The magnitude of the electrostatic force is now |Fe| = k q1q2/r2
=> |Fe|= 9 x 109 x (0.3 x 10-6 )(0.3 x 10-6)/(3.3 x 10-2)2 = 0.74 N
Since both spheres have a negative charge, the electrostatic force is repulsive.

Question 3] A small metal sphere (B) with a negative charge of 2.0 x 10-6 C is placed midway between two similar spheres (A and C) with positive charges of 1.5 x 10-6 C that are 3.0 cm apart (see Figure). Use a vector diagram to find the net electrostatic force acting on sphere B.

Question 4] A small metal sphere (B) with a negative charge of 2.10 x 10-6 C is placed midway between two similar spheres (A and C) 3.30 cm apart with positive charges of 1.00 x 10-6 C and 1.50 x 10-6 C, respectively, as shown in Figure. If the three charges are along the same line, calculate the net electrostatic force on the negative charge.

Question 5] A small metal sphere A with a negative charge of 2.10 x 10-6 C is 2.00 x 10-2 m to the left of another similar sphere B with a positive charge of 1.50 x 10-6 C. A third sphere C with a positive charge of 1.80x 10-6 C is situated 3.00 x 10-2 m directly below sphere B (Figure). Calculate the net electrostatic force on sphere B.

Question 6] A small metal sphere A with a charge of -2.10 x 10-6 C is 2.00 x 10-2 m to the left of a second sphere B with a charge of +1.50 x 10-6 C.
A third sphere C with a charge of +1.80 x 10-6 C is situated 3.00 x 10-2 m directly below sphere B. Calculate the net electrostatic force on sphere C.

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