Question

A + 6 nC point charge and a - 2.64 nC point charge are 6.79 cm apart. What is the electric field strength at the midpoint between the two charges?

Answer 1

The given problem can be exemplified in the following diagram:

The electric field of a positive charge points outwards from the charge and the electric field of negative charge points toward the charge.

The total electric field is then the sum of both electric fields since they point in the same direction:

`E=E_6+E_(2.64)`

The formula for the electric field is given by:

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

Now, we substitute in the formula for the total electric field:

`E=k(q_6)/(r^2)+k(q_(2.64))/(r^2)`

Taking the constant "k" and "r" as common factors:

`E=(k)/(r^2)(q_6+q_(2.64))`

The radius "r" is the mid-point between the charges, therefore, it is:

`r=(6.79cm)/(2)=3.4cm`

Now, we substitute the values:

`E=((9*10^9(Nm^2)/(C^2)))/((0.034m)^2)(6*10^(-9)C+2.64*10^(-9)C)`

Solving the operations:

`E=67266.44(J)/(C)`