Respuesta :
The magnitude and direction of the electric field is equal to 3.45 x10^13 N/m and the direction in towards the negative charge.
Electric field
To calculate the electric field of a point it is necessary to use the following expression:
[tex]E = k\frac{Q}{d^{2}}[/tex]
Thus, calculating the field formed by each charge we have:
- –8.0 µC
[tex]E = 9 \times 10^{9} \times \frac{8}{36\times 0.0001}[/tex]
[tex]E = 2\times 10^{13} V/m[/tex]
- +5.8 µC
[tex]E = 9 \times 10^{9} \times \frac{5.8}{36\times 0.0001}[/tex]
[tex]E = 1.45 \times 10^{13} N/m[/tex]
The direction of the electric field of the negative charge is towards the charge (attracted) and of the positive charge is repelled.
So, we can add the two values of field found, since the two are directed towards the negative charge, so that :
[tex]E = 2 \times 10^{13} + 1.45 \times 10^{13}[/tex]
[tex]E = 3.45 \times 10^{13} N/m[/tex]
So, the magnitude of eletric field is equal to 3.45 x10^13 N/m and the direction in towards the negative charge.
Learn more about electric field in: brainly.com/question/26446532
