Question

The electrostatic force between two charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum is $F$. If a dielectric medium of dielectric constant $K$ is introduced between them, the new force is

• A. $F / K$
• B. $K F$
• C. $F / K^2$
• D. $K^2 F$

Hint:

Remember that introducing a dielectric medium with dielectric constant \( K \) reduces the electrostatic force by a factor of \( K \).

Answer 1

The Correct Option is A

Step 1: Recall Coulomb's law.
The force between two charges in vacuum is \[ F = k\frac{q_1 q_2}{r^2} \] where $k$ is Coulomb's constant.

Step 2: See what a dielectric does.
A dielectric is an insulating material. When placed between the charges, it gets slightly polarized and weakens the field. This makes the force smaller.

Step 3: Write the new force.
Inside a medium with dielectric constant $K$, the force becomes \[ F' = k\frac{q_1 q_2}{K r^2} \]

Step 4: Compare with the original.
Since the original force is $F = k\dfrac{q_1 q_2}{r^2}$, we can write \[ F' = \frac{F}{K} \]

Step 5: Understand the meaning.
The dielectric constant $K$ is more than one, so dividing by $K$ makes the force smaller. The medium shields the charges from each other.

Step 6: State the answer.
The new force is the old force divided by $K$. \[ \boxed{F' = \frac{F}{K}} \]