Q4 2pts Two point charges lie on the x-axis. A charge of +6.2 ?C is at the origi

Question

Q4 2pts Two point charges lie on the x-axis. A charge of +6.2 ?C is at the origin, and a charge of-9.5 ?C is at 10.0cm. What is the net electric field? At: a) x=-4.0cm b) x +4.0cm 41 912 -4.0 cm 0 4.0 cm 10.0 cm Use Electric field equation to find the magnitude and direction of the electric fields created by each of the two charges at the specified locations. Then find the vector sum of those fields to find the net electric field. Atx-4.0 cm the field from qi will point in the direction and the field from q2 will point in the direction.

Explanation / Answer

E = Net Electric Field

E = Ke((q1/r1^2)+(q2/r2^2))

a) at x = - 4 cm

Ke = 8.99*10^9 N.m 2 /C2

q1 = -6.2 ?C

q2 = 9.5 ?C

r1 = 0.040 m

r2= 0.1+0.04 = 0.14 m

E = Ke((q1/r1^2)+(q2/r2^2))

= (8.99*10^9)*(((-6.2*10^-6)/0.040)+((9.5*10^-6)/0.14))

= -783414.29

b) at x = + 4 cm

E = Ke((q1/r1^2)+(q2/r2^2))

Ke = 8.99*10^9 N.m 2 / C2

q1 = +6.2 ?C

q2 = +9.5 ?C

r1 = 0.040 m

r2= 0.1-0.04 = 0.06 m

E = Ke((q1/r1^2)+(q2/r2^2))

= (8.99*10^9)*(((6.2*10^-6)/0.040)+((9.5*10^-6)/0.06))

= 2816866.67

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