An electric field of 10^5 N//C points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of +2muC and -5muC at this spot?
Answers
Answer:
F = q Ē
= 2 × 10^-6 × 10^5
= 0.2 N due west
F = 5 × 10^-6 × 10^5
= 0.5 due east
PLEASE MARK AS BRAINLIEST.
The magnitude and direction of force that acts on the charges +2*10^-6 C and -5*10^-6 C are 0.2 N in west and 0.5 N in East. Now
1. We know that Force in an electrostatic field= q*E
2.Where q= charge released in the field and E = field strength.
3. Now for +2*10^-6 C charge:
Force= q*E
= +2*10^-6*10^5
=0.2 N
4. Since the charge is positive so it will experience force in the West.
5. Now for -5*10*-6 C charge:
Force=q*E
= 5*10^-6*10^5
= 0.5 N (Magnitude only)
5. Since charge is negative . Therefore it will experience force in the East.