Question

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

Answer 1

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 E_net = 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 E_net has an negative sign.

Therefore, E_net = 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 E_net should have negative sign. Therefore, E_net = -0.76 x 10⁸ N/C.