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

2.37 N.

Step-by-step 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 = (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 = \frac 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* 10^(-6)\ C.`

Therefore, the electric force at that point is given by

`\rm F = 3.00* 10^(-6)* 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.