Question

A charge of 7. 2 × 10-5 C is placed in an electric field with a strength of 4. 8 × 105StartFraction N over C EndFraction. If the electric potential energy of the charge is 75 J, what is the distance between the charge and the source of the electric field? Round your answer to the nearest tenth. M.

Answer 1

The distance between the charge and the source of the electric field is 2.2 m.

Relationship between potential energy and Distance:

The distance of charge from the source is directly proportional to the potential energy.

$r = \dfrac {U}{Eq}$

Where,

$r$ - distance

$U$- potential energy =  75 J

$E$ - Electric field = $4. 8 \times 10^5 \rm \ N/C$

$q$ - charge = $7. 2 \times 10^{-5} \rm \ C$

Put the values in the formula,

$r = \dfrac {75}{ 4. 8 \times 10^5 \rm \ N/C\times 7. 2 \times 10^{-5} \rm \ C}\\\\ r = 2.2 \rm \ m$

Therefore, the distance between the charge and the source of the electric field is 2.2 m.

Learn more about the Electric field:

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