Question

1. Imagine that the magnitude of two point charges is halved and the distance between them is also halved. The force of interaction between them will be

(a) doubled. (d) halved

(c) be tripled

(d) same​

Answer 1

Answer:

(d) same   is the correct answer.

Explanation:

According to coulomb's law,

Force between two charges is given by = $\frac{1}{4\pi \epsilon_o} \frac{q_1 q_2}{r^2}$

where, $q_1, q_2$ are magnitude of the two charges respectively.

and $r$  is the distance between the charges.

If magnitude of the two charges is halved, then

their magnitudes become $\frac{q_1}{2} , \frac{q_2}{2}$  respectively.

and distance between them is also halved.

distance between them becomes $\frac{r}{2}$.

New force of interaction is = $\frac{1}{4\pi \epsilon_o} \frac{\frac{q_1}{2} .\frac{q_2}{2} }{(\frac{r}{2})^2 } = \frac{1}{4\pi \epsilon_o} \frac{q_1 q_2}{r^2}$

So, the new force of interaction is equal in magnitude as before.

Therefore, option (d) is correct.

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Answer 2

Concept Introduction: Charges follow Coulomb's Law.

Given:

We have been Given: Charge of two point charges is halved and distance between them is also halved.

To Find:

We have to Find: The force of interaction between them will be:

1. doubled.
2. halved
3. be tripled
4. same

Solution:

According to the problem, Magnitude of the charges are,

$\frac{q1}{2} \: \frac{q2}{2}$

Distance separating them is,

$\frac{r}{2}$

therefore following Coulomb's Law, i.e.,

$f = \frac{1}{4\pi \times e} \times \frac{q1 \times q2}{ {r}^{2} }$

therefore putting the values of the quantities,

$f = \frac{1}{4\pi \times e} \times \frac{ \frac{q1}{2} \times \frac{q2}{2} }{ (\frac{r}{2} )^{2}}$

that will give us

$f = \frac{1}{4\pi \times e} \times \frac{4 \times q1 \times q2}{4 {r}^{2} } = \frac{1}{4\pi \times e} \times \frac{q1 \times q2}{ {r}^{2} }$

therefore the force remains same. So the correct option is Option No. 4

Final Answer: The Force of interaction between the charges will be Same.

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