**Electric Field due to Multiple Point Charges**

Part A = 0,0.300 N/C

Part B = 0.300 nC

**Solutions Below:**

_{1}= 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q

_{2}= 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.

Part A

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures.

This is a long problem, but I promise it’s simple if you break it down into parts and solve step by step. Make sure to watch the units since the problem wants the answer in nano Coulombs

The first thing to do is recognize that this is a vector type problem that requires you to set up a triangle in order to solve for the x and y components from each charge. Start by finding the lengths of each side of the triangle.

In the picture below, the side in red (from q_{2} to point A) has a length of 15m (use the Pythagorean theorem, e.g. sqrt(x^{2} + y^{2}) = sqrt((-9)^{2} + 12

Next, find the angles formed by each side. The angle formed by q_{1} and the x-axis is 36.87° (sin(θ) = 12m / 20m, so sin^{-1}(12/16) = 36.87°). The angle formed by q_{2} is 53.13°

Now find the electric field from each charge at Point A, using the formula F = kq / r^{2}. Note that ‘k’ is the Coulomb Constant, 8.9875 * 10^{9} Nm^{2}/C^{2}. Also remember that the charges were given in nC (nano Coulombs, which is to the -9th power):

The electric field from q_{1} is given by:

F_{1} = kq_{1} / r1^{2}

F_{1} = 8.9875 * 10^{9} * (8.00 * 10^{-9}) / 20^{2}

F_{1} = 0.18 N/C

Likewise, for q_{2}:

F_{2} = kq_{2} / r2^{2}

F_{2} = 8.9875 * 10^{9} * (6.00 * 10^{-9}) / 15^{2}

F_{2} = 0.24 N/C

To find the x and y components, just use trigonometry (note that the direction of the field is away from the charges, since the charges are positive, so q_{1}‘s x component will be negative, q_{2}‘s x component will be positive, and both y components will be positive:

F_{1, x} = F_{1} * cos(θ_{1})

F_{1, x} = 0.18 * -cos(36.87)

F_{1, x} = -0.144 N/C

F_{1, y} = F_{1} * sin(θ_{1})

F_{1, y} = 0.18 * sin(36.87)

F_{1, y} = 0.108 N/C

F_{2, x} = F_{2} * cos(θ_{2})

F_{2, x} = 0.24 * cos(53.13)

F_{2, x} = 0.144 N/C

F_{2, y} = F_{2} * sin(θ_{2})

F_{2, y} = 0.24 * sin(53.13)

F_{2, y} = 0.192 N/C

So the sum of the x components is 0 (+0.144 from q_{2}, -0.144 from q_{1}. The sum of the y components is 0.300 (0.144 + 0.192):

0,0.300 N/C

Part B

_{3}is now placed at point B, located at coordinates (0 m, 15.0 m).

Find the magnitude and sign of q

_{3}needed to make the total electric field at point A equal to zero.

Express your answer in nanocoulombs to three significant figures.

This is kind of a tricky question because if you just added the *total* fields from Part A (0.18 and 0.24), you wouldn’t get the right answer. What you have to do is add the x and y components individually, in case they cancel each other out (partially or totally). In Part A you saw that the x-components canceled each other entirely and only the y components remained. Find the overall electric field at Point A by (vector math):

F_{A} = sqrt((F_{q1, x} + F_{q2, x})^{2} + (F_{q1, y} + F_{q2, y})^{2})

F_{A} = sqrt((0.144 – 0.144)^{2} + (0.108 + 0.192)^{2}))

F_{A} = sqrt(0 + 0.09)

F_{A} = 0.3 N/C

Now you can find the charge at Point B needed to zero out the electric field at Point A. Since the electric field at Point A is 0.3 N/C, we need a charge on Point B that will give us this same field value, but in the opposite direction. Since q_{1} and q_{2} were both positive, the charge at Point B should also be positive (to push point A back down):

Distance (“r”) between Point B and Point A:

r = 15m – 12m

r = 3m

F = kq_{B} / r^{2}

0.3 = 8.9875 * 10^{9} * q_{B} / 3^{2}

0.3 = 8.9875 * 10^{9} * q_{B} / 9

q_{B} = 3.00 * 10^{-10} C

Since Mastering Physics asks for the answer in nano Coulumbs (-9th power):

0.300 nC

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