Question

1.Two point charges, q₁= +6 nC and q₂= +5 nC, are located on the x-axis at x= 0 and x=5.00 m. What is the electric field at point x=1.27 m? I got 33.4801 is ths correct?

2.Two point charges, q₁= +2 nC and q₂= -4 nC, are located on the x-axis at x= 0 and x=5.00 m. What is the electric field at point x=5 m? I got is this correct .7199?

3.Two point charges, q₁= +4 nC and q₂= -14 nC, are located on the x-axis at x= 0 and x=8 m. What is the distance of ZERO electric field from the positive charge?

Answer 1

1.

Enet = E1 - E2 = k*q1/r1^2 - k*q2/r2^2

r1 = 1.27 m & r2 = 3.73 m

So,

Enet = 9*10^9*6*10^-9/1.27^2 - 9*10^9*5*10^-9/3.73^2 = 30.24 N/C

2.

Now here due to q2, there will be no electric field since r = 0

So, net electric field will be

Enet = E1 = k*q1/r1^2

Enet = 9*10^9*2*10^-9/5^2 = 0.72 N/C (So yes your answer is correct.)

3.

Since direction of electric field is away from positive charge and towards the negative charge, So electric field cannot be zero between both charges, It will be zero towards the left of origin, since |q1| < |q2|.

Suppose at a point P(distance d from origin towards the left of origin), total electric field is zero.

Ep = E1 + E2 = 0

E1 = -E2

kq1/d^2 = kq2/(d+ x)^2

[d/(d + x)]^2 = q1/q2

[d/(8 + d)]^2 = 4/14 = 2/7

d/(8 + d) = sqrt (2/7) = 0.5345

d = 0.5345*8/(1 - 0.5345) = 9.18 m

d = 9.18 m to the left side of positive charge. (at x = -9.18 m)

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