search

mMCQ.

Navigation Menu

Step into mMCQ.

Launch mMCQ. diagnostic

Explore mMCQ.

MDCAT prepFree DiagnosticPricing & SubscribeSign in

Resources

Terms & Conditions

mMCQ.

© 2021 - 2025 mMCQ.All rights reserved.

  1. Home
  2. MDCAT
  3. Physics
  4. Electrostatics
MDCAT
Physics
2021

Physics · Electrostatics

Work through this past-paper style MCQ, then read the full explanation. Practice more physics questions on mMCQ with adaptive drills and topic analytics.

Question

If the distance between two point charges is halved, the force between them would be ___.

Options
  • A

    Half

  • B

    Double

  • C

    One fourth

  • D

    Four times

Explanation

The force between two point charges is given by Coulomb's law:

F = (kq1q2)/r2

where F is the force, q1 and q2 are the charges, r is the distance between them, and k is the Coulomb constant.

If the distance between the two charges is halved (i.e., r is divided by 2), then the force between them becomes:

F' = (kq1q2)/(r/2)2

F' = (kq1q2)/(r2/4)

F' = 4xkq1q2)/r2 

F' = 4F

Thus, the force between the two charges is quadrupled (i.e., becomes four times as much) if the distance between them is halved

Incorrect as per formulaThe force between two point charges is given by Coulomb's law:

F = (kq1q2)/r2

where F is the force, q1 and q2 are the charges, r is the distance between them, and k is the Coulomb constant.

If the distance between the two charges is halved (i.e., r is divided by 2), then the force between them becomes:

F' = (kq1q2)/(r/2)2

F' = (kq1q2)/(r2/4)

F' = 4x(kq1q2)/r2

F' = 4F

Thus, the force between the two charges is quadrupled (i.e., becomes four times as much) if the distance between them is halved.

Incorrect as per formulaThe force between two point charges is given by Coulomb's law:

F = (kq1q2)/r2

where F is the force, q1 and q2 are the charges, r is the distance between them, and k is the Coulomb constant.

If the distance between the two charges is halved (i.e., r is divided by 2), then the force between them becomes:

F' = (kq1q2)/(r/2)2

F' = (kq1q2)/(r2/4)

F' = 4x(kq1q2)/r2

F' = 4F

Thus, the force between the two charges is quadrupled (i.e., becomes four times as much) if the distance between them is halved.

Incorrect as per formulaThe force between two point charges is given by Coulomb's law:

F = (kq1q2)/r2

where F is the force, q1 and q2 are the charges, r is the distance between them, and k is the Coulomb constant.

If the distance between the two charges is halved (i.e., r is divided by 2), then the force between them becomes:

F' = (kq1q2)/(r/2)2

F' = (kq1q2)/(r2/4)

F' = 4x(kq1q2)/r2

F' = 4F

Thus, the force between the two charges is quadrupled (i.e., becomes four times as much) if the distance between them is halved.

The force between two point charges is given by Coulomb's law:

F = (kq1q2)/r2

where F is the force, q1 and q2 are the charges, r is the distance between them, and k is the Coulomb constant.

If the distance between the two charges is halved (i.e., r is divided by 2), then the force between them becomes:

F' = (kq1q2)/(r/2)2

F' = (kq1q2)/(r2/4)

F' = 4x(kq1q2)/r2 

F' = 4F

Thus, the force between the two charges is quadrupled (i.e., becomes four times as much) if the distance between them is halved.

Keep practising Physics

Take the free 2 minutes diagnostic to map weak topics, then unlock full Biology, Chemistry, Physics, English, and Reasoning practice for PKR 15,000/month.

Start free diagnosticExplore MDCAT

Tagged under Physics · Electrostatics · 2021