Question

Two point charges of +2.00 μC and -2.00 μC are 0.100 m apart. What is the electric field at the point midway between the two charges?Group of answer choices2.88 x 107 N/C1.44 x 107 N/C7.19 x 106 N/C0 N/C

Answer 1

The electric field at the point midway between the two charges is 7.19 x 106 N/C.

Step-by-step explanation:

To find the electric field at the point midway between two charges, we can use the formula:

E = k(Q1-Q2)/(2r^2)

where E is the electric field, k is the electrostatic constant, Q1 and Q2 are the charges, and r is the distance between the charges.

In this case, Q1 and Q2 are +2.00 μC and -2.00 μC, respectively, and the distance between them is 0.100 m.

Substituting these values into the formula, we get:

E = (8.99 x 10^9 N m^2/C^2) * (2.00 x 10^-6 C - (-2.00 x 10^-6 C))/(2 * 0.100 m^2)

E = 7.19 x 106 N/C

Therefore, the electric field at the point midway between the two charges is 7.19 x 106 N/C.

Answer 2

Take into account that the magnitude of electric field at point P, produced by a charge q is given by:

`E=k(q)/(r^2)`

where k (9*10^9 N/m^2C^2) is the Coulomb's constant, r is the distance from the charge to the point P and q is the magnitude of the charge.

An illustration of the given situation is shown below:

As you can notice, at point P the electric fields generated by each charge point to the negative charge. The signs of the charge determine the direction of the electric field.

The magnitude of each electric field is the same, due to the distance to each charge is the same and the magnitude of the charge is equal. Then, The magnitude of the total electric field is:

`\begin{gathered} E=E_1+E_2=2(9\cdot10^9(N)/(m^2C^2))((2.00\cdot10^(-6)C)/((0.05m)^2)) \\ E\approx1.44\cdot10^7(N)/(C) \end{gathered}`

Hence, the electric field at a point midway the given charges is approximately 1.44*10^7N/C