The electrostatic force ( F ) between two plates of a charged capacitor is directly proportional to the product of the charges ( $q_ 1$ and $q_2$ ) on each plate and inversely proportional to the square of the distance ( d ) between the plates.
A) $F = \frac { q _ { 1 } } { d q _ { 2 } }$
B) $F = \frac { d } { q 1 q 2 }$
C) $F = \frac { q 1 q 2 } { d }$
D) $F = \frac { q 1 q 2 } { d ^ { 2 } }$

Question

Quizplus · Accepted Answer

According to the statement of the problem, the electrostatic force (F) is directly proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (d) between the plates. The exact mathematical relationship can be written as:

F ∝ (q1q2)/d^2

where the symbol ∝ means "is proportional to". To remove the proportionality sign and obtain an exact equality, we need to introduce a proportionality constant, which is usually denoted by k:

F = k(q1q2)/d^2

The value of the constant k depends on the units used to measure F, q1, q2, and d, as well as on the properties of the medium between the plates (e.g. the permittivity of free space). However, we don't need to know the exact value of k to answer the question. We just need to find the expression that correctly represents the relationship between F, q1, q2, and d. By comparing the given options with the above equation, we can identify that only option D matches:

F = (q1q2)/d^2

Therefore, the answer is D, and the explanation is as above.

The Electrostatic Force ( F ) Between Two Plates of a Charged