Question

Consider two positive charges separated by a distance ((A 20)/100) meters. One charge has a value of (B 5) mC (milli-coulombs) and is located to the right of the other (C 5) mC charge. Assume the left charge is the origin and the other charge is located with a positive horizontal position. How large and in what direction is the electrostatic field at the mid-point, directly between the two charges

Answer 1

Answer:

E_total = 0

Explanation:

In this exercise we are asked for the electric field created by two charges at the midpoint between them, since the electric field is a vector magnitude it must be added as vectors. The equation that describes the electric field for a charge is

           $E = k \frac{q}{r^{2} }$

in this case the two charges are positive, therefore the field is salient, in the adjoint we can see a diagram of the system

          E_total = E₁ -E₂

we calculate the electric field of charge 1 with q = + 5 10⁻³ C and a distance r = L / 2 = 100/2 = 50 m

           E₁ = $\frac{9 10^{9} \ 5 10^{-3} }{ 50^{2} }$

           E₁ = 1.8 10⁴ N / C

the magnitude of the electric field of charge 2 is equal, but the sense is opposite.

Since the two charges have the same sign, the addition of vectors gives zero

           E_total = 0

if the charges were of different signs, the fields would be added

            E_total = 3.6 10⁴ N / A