Three charges are situated as shown in the diagram (each grid line is separated

Question

Three charges are situated as shown in the diagram (each grid line is separated by 1 meter). Two of the charges at (-2,2) and (2,-2) (the upper left and lower right) have a charge of -3 µC (-3×10-6 C) but the upper right charge (2,2) is +4 µC (+4×10-6 C) .

1) What is the magnitude of the net electric field at the point (0, 0)?

a) 2,250 N/C

b) 4,500 N/C

c) 9,000 N/C

d) 11,250 N/C

e) 0 N/C

3) What is the net electric potential at the point (-2, -2)?

a) -2,250 V

b) -4,500 V

c) -7,136 V

d) -19,864 V

e) 0 V

6) What is the magnitude of the net force on the upper right charge?

a) 0 N

b) 0.0035 N

c) 0.0055 N

d) 0.0075 N

e) 0.0095 N

Explanation / Answer

at (0,0) electric field due to charge on (-2,2) and (2,-2) will cancel each other
So, net electric field will be due to only charge at (2,2)
E = K*q/r^2
q = 4*10^-6 C
r = sqrt (2^2 + 2^2) = sqrt(8)
E = K*q/r^2
= (9*10^9)*(4*10^-6) / 8
= 4500 N/C
Answer: b

q1 = -3×10-6 C
q2 = -3×10-6 C
q3 = 4×10-6 C
r1 = 4 m
r2 = 4 m
r3 = sqrt(4^2+4^2) = 5.66 m

V = V1 + V2 + V3
= K*(q1/r1 + q2/r2 + q3/r3)
=(9*10^9)*((-3*10-6)/4 -(3*10-6)/4 +(4×10-6)/5.66)
=(9*10^9)*(-1.5*10^-6 + 7.07*10^-7)
= -7136 V
  
Answer: c

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