Consider the three point charges shown below. • • • 5C 4C 3C If the distance
ID: 1508922 • Letter: C
Question
Consider the three point charges shown below.
• • •
5C 4C 3C
If the distance between the 5 C charge and the 4 C charge is 0.25m, and the distance between the 4 C charge and the 3 C charge is 0.15m.
What is the total force felt by the 4 C charge?
Be sure to include the magnitude and direction of the force.
What is the electric field at the position of the 4 C charge?
Once again be sure to include the magnitude and direction since electric field is a vector.
Answer provided is
F = 7.68 x 10 12 N to the left, E = 1.92 x 10^12 N/C to the right
I need to know how to get to the answer.
Explanation / Answer
Alike charge repels.
hence force on 4C due to 5C will be to the right
and due to 3C will be to the left.
and magnitude of force between two charges, F = k q1 q2 /d
Fnet = [ (9 x 10^9)(5)(4) / (0.25^2) ] - [ (9 x 10^9) (4)(3) / (0.15^2) ]
= (2.88 x 10^12 ) - ( 4.8 x 10^12 )
= - 1.92 x 10^12 N ( minus sign means to the left )
{F = 7.68 x 10^12 and E = 1.92 x 10^12 N/C will be when 4C and 3C are negative charge }
then Fnet = - (2.88 x 10^12 ) - ( 4.8 x 10^12 ) = - 7.68 x 10^12 N
and F = q E
- 7.68 x 10^12 = ( - 4 ) E
E = 1.92 x 10^12 N/C to the right