Question

An electric field of 790,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -3.00 uC at this spot? (14C = 10 6C) Give your answer in Sl unit rounded to two decimal places.

Answer 1

Answer:

2.37 N.

Explanation:

According to Coulomb's law, the magnitude of electrostatic force of interaction between two static point charges $\rm q_1$ and $\rm q_2$, separated by a distance $\rm r$, is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where, $k$ = Coulomb's constant.

The direction of this force is along the line joining the two charges, from positive to negative charge.

The electric field at a point is defined as the amount of electrostatic force experienced per unit small positive test charge, placed at that point.

Therefore,

$\rm E = \dfrac Fq\\\\\Rightarrow F = qE.$

We have,

• Electric field, $\rm E = 790,000\ N/C.$
• Charge placed at given spot, $\rm q=-3.00\ \mu C =-3.00\times 10^{-6}\ C.$

Therefore, the electric force at that point is given by

$\rm F = 3.00\times 10^{-6}\times 790,000=2.37\ N.$

The negative sign indicates that the force is attractive in nature.

Thus, the magnitude of this force = 2.37 N.