Question: The force experienced by a charge placed in an electric field of strength E is given by: (A) \(F =...

Question

The force experienced by a charge placed in an electric field of strength E is given by:
(A) $F = \dfrac{q}{E}$
(B) $F = \dfrac{E}{q}$
(C) $F = qE$
(D) $F = E - q$

Answer 1

Solution

When a test charge is brought in the vicinity of another charge then, the test charge experiences a force due to the other charged particle. Similarly, the other charge also experiences the same amount of force due to the test charge but in the opposite direction. The force is equal and opposite.Depending upon the test charge the force changes. Hence, we define force as the product of test charge with some parameter known as electric field which is independent of test charge.

Complete step by step answer:
The electrostatic force is directly proportional to the product of charges q and Q placed in the vicinity of each other and inversely proportional to the square of the distance between them. This is the Coulomb’ Law. It is purely experimental formula i.e. it was derived based on the data recovered from an experiment. Mathematically,
$F = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{{Qq}}{{{r^2}}}$
Now, the electric field is described as the physical field which surrounds the static electric charge. It is a real parameter and not some imaginary parameter to ease our solving of problems like potential.
The electric field is defined for every point in the space (depends upon r). When a test charge q is placed at that point it will experience a force depending upon the amount of charge it carries i.e, q.
So, the electric field is independent of test charge q but is inversely proportional to the square of the distance. Mathematically,
$E = \dfrac{1}{{4\pi {\varepsilon _o}}}\dfrac{Q}{{{r^2}}}$
Therefore,

\dfrac{F}{E} = q \\\ F = qE \\\

Hence, the correct answer is option C.

Note: The expressions used here have been used in scalar form. Vector form could also have been used but since we needed to find the electric field intensity in terms of force exerted between the charged particles, the relation would have come out the same (but in vector form which was not required).