Question

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?

Answer 1

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.

Answer 2

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.