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Electric Field Due to Two Charges what is the the electric field at a point midway...

Electric Field Due to Two Charges what is the the electric field at a point midway between a-8.46 μC and a +5.42 μC charge 3.87 cm apart? Take the direction towards the positive charge to be positive. Submit Answer Incorrect. Tries 7/10 Previous Tries

Electric Field Due to Two Charges what is the the electric field at a point midway between a-8.46 μC and a +5.42 μC charge 3.87 cm apart? Take the direction towards the positive charge to be positive. Submit Answer Incorrect. Tries 7/10 Previous Tries
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Answer:

Consider two opposite charges placed at a distance d, here I assumed the negative charge on the left side and the positive on the right side.

Given that, q1 = -8.6 x 10-6 C, q2 = +5.42 x 10-6 C and distance of separation d = 3.87 cm = 0.0387 m and d/2 = 0.01935

The electric file due to the charge q1 at a point midway between them is E1 = k (-8.6 x 10-6 C)./ (0.01935 m)2 = -2.06 x 108 N/C and its direction is pointing left because E fields point in towards negative charges. That means towards left side.

since k = 9 x 109 N m2/C2

and the electric field due to the charge q2 at the mid point is E2 = k (+5.42 x 10-6 C) / (0.01935 m)2 = +1.30 x 108 N/C and its direction pointing towards left because E fields point away from positive charges. That means E2 towards left side.

Therefore, the net electric field at the mid point is Enet = E1 + E2.

From the given question, we need to consider that the direction towards the positive charge to be positive. That means the direction towards the negative charge will be negative. So the right side will be negative.

Using the sign convention given in the question, the net field's direction pointing towards right side, that means Enet has an negative sign.

Therefore, Enet = E1 + E2 = -2.06 x 108 N/C + 1.30 x 108 N/C = 0.76 x 108 N/C and its direction is pointing towards left or towards negative charge, so Enet should have negative sign. Therefore, Enet = -0.76 x 108 N/C.

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