SATHEE: Electrostatics Question 193

Electrostatics Question 193

Question: Two point charges $+ 3 μ C$ and $+ 8 μ C$ repel each other with a force of $40 N$ . If a charge of $- 5 μ C$ is added to each of them, then the force between them will become [SCRA 1998; JIPMER 2000]

Options:

A) $- 10 N$

B) $+ 10 N$

C) $+ 20 N$

D) $- 20 N$

Show Answer

Answer:

Correct Answer: A

Solution:

In second case, charges will be $- 2 μ C$ and $+ 3 μ C$

Since $F \propto Q_{1} Q_{2}$ i.e. $\frac{F}{F^{'}} = \frac{Q_{1} Q_{2}}{Q_{1}^{'} Q_{2}^{'}}$ $\frac{40}{F^{'}} = \frac{3 \times 8}{- 2 \times 3} = - 4$

therefore $F^{'} = 10 N$ (Attractive)

Electrostatics Question 193

Question: Two point charges $+ 3 μ C$ and $+ 8 μ C$ repel each other with a force of $40 N$ . If a charge of $- 5 μ C$ is added to each of them, then the force between them will become [SCRA 1998; JIPMER 2000]

Options:

Answer:

Solution:

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Electrostatics Question 193

Question: Two point charges +3μC and +8μC repel each other with a force of 40N . If a charge of −5μC is added to each of them, then the force between them will become [SCRA 1998; JIPMER 2000]

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: Two point charges $+ 3 μ C$ and $+ 8 μ C$ repel each other with a force of $40 N$ . If a charge of $- 5 μ C$ is added to each of them, then the force between them will become [SCRA 1998; JIPMER 2000]