Question

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m

PART A

Find the electric field (magnitude and direction) at point on the x-axis at x = 0.200 m.

PART B

Find the electric field (magnitude and direction) at point on the x-axis at x = 1.20 m.

PART C

Find the electric field (magnitude and direction) at point on the x-axis at x = -0.200 m.

Answer 1

electric field is given by

E = k*Q/R^2

q1 = -4.00 nC at origin & q2 = -5.50 nC at x = 0.800 m

Remember that direction of electric field is away from positive charge and towards the negative charge. So

Part A. Net electric field x = 0.200 m will be: (Point x is between both charges)

Electric field due to q1 at x will be towards the charge in -ve x-axis, since q1 is negative

Electric field due to q2 at x will be towards the charge in +ve x-axis, since q2 is negative

So net electric field at x will be:

Enet = -E1 + E2

Enet = -k*q1/d1^2 + k*q2/d2^2

d1 = distance between q1 and x = 0.200 m

d2 = 0.800 - 0.200 = 0.600 m

q1 = -4.00 nC = -4.00*10^-9 C

q2 = -5.50 nC = -5.50*10^-9 C

So Using these values:

Enet = -k*[q1/d1^2 - q2/d2^2]

Enet = -9*10^9*[4.00*10^-9/0.200^2 - 5.50*10^-9/0.600^2]

Enet = -762.5 N (Use this if you have to use both magnitude and direction together, here -ve sign shows that electric field is towards -ve x-axis)

If you've two separate place magnitude and direction, then use answer given below

Magnitude of electric field = |Enet| = 762.5 N

(-ve sign means net electric field will be towards left or -ve x-axis)

Part B.

Net electric field x = 1.200 m will be: (Point x is on the right of both charges)

Electric field due to q1 at x will be towards the charge in -ve x-axis, since q1 is negative

Electric field due to q2 at x will be towards the charge in -ve x-axis, since q2 is negative

So net electric field at x will be:

Enet = -E1 - E2

Enet = -k*q1/d1^2 - k*q2/d2^2

d1 = distance between q1 and x = 1.200 m

d2 = 1.200 - 0.800 = 0.400 m

q1 = -4.00 nC = -4.00*10^-9 C

q2 = -5.50 nC = -5.50*10^-9 C

So Using these values:

Enet = -k*[q1/d1^2 + q2/d2^2]

Enet = -9*10^9*[4.00*10^-9/1.200^2 + 5.50*10^-9/0.400^2]

Enet = -334.4 N (Use this if you have to use both magnitude and direction together, here -ve sign shows that electric field is towards -ve x-axis)

If you've two separate place magnitude and direction, then use answer given below

Magnitude of electric field = |Enet| = 334.4 N

(-ve sign means net electric field will be towards left or -ve x-axis)

Part C.

Net electric field x = -0.200 m will be: (Point x is on the left of both charges)

Electric field due to q1 at x will be towards the charge in +ve x-axis, since q1 is negative

Electric field due to q2 at x will be towards the charge in +ve x-axis, since q2 is negative

So net electric field at x will be:

Enet = E1 + E2

Enet = k*q1/d1^2 + k*q2/d2^2

d1 = distance between q1 and x = 0.200 m

d2 = 0.800 - (-0.200) = 1.000 m

q1 = -4.00 nC = -4.00*10^-9 C

q2 = -5.50 nC = -5.50*10^-9 C

So Using these values:

Enet = k*[q1/d1^2 + q2/d2^2]

Enet = 9*10^9*[4.00*10^-9/0.200^2 + 5.50*10^-9/1.000^2]

Enet = 949.5 N (Use this if you have to use both magnitude and direction together, here +ve sign shows that electric field is towards +ve x-axis)

If you've two separate place magnitude and direction, then use answer given below

Magnitude of electric field = |Enet| = 949.5 N

(+ve sign means net electric field will be towards right or +ve x-axis)

Let me know if you've any query.