Respuesta :
The electric field due to charge q1 and q2 at the middle point of the line joining the charges q1 and q2 is 8.64*10^(4) N/C.
Note: You have not given the point where the resultant magnetic field has to be calculated. Most probably it is asked to find the electric field at the middle point of the line joining the point charges.
Electric field: The electric force exerted on a unit charge is called the electric field. Electric field due to a charge is calculated using the formula,
E=kq/r^2
where k is a constant whose value is 9*10^(9) N m^2/ C^(2), q is the charge and r is the distance from the charge to the point where the electric field has to be calculated. In the given case, the electric field has to be calculated at the middle point of the line joining charges q1 and q2.
Calculation of electric field due to charge q1:
Given that q1=+12 nC or q1=+12*10^(-9) C and the distance of the charge q1 from the center d1=10.0/2 cm or d1=5*10^(-2) m, the magnitude of the electric field E1 due to charge q1 is,
E1=kq1 /d1^(2)
E1=9*10^(9)*12*10^(-9) / (5*10^(-2))^2
E1=4.32*10^(4) N/C
The direction of the electric field E1 at the middle point is towards the negative charge.
Calculation of electric field due to charge q2:
Given that q2=-12 nC or q1=-12*10^(-9) C and the distance of the charge q2 from the center d2=10.0/2 cm or d2=5*10^(-2) m, the magnitude of the electric field E2 due to charge q2 is,
E2=kq2 /d2^(2)
E2=9*10^(9)*12*10^(-9) / (5*10^(-2))^2
E2=4.32*10^(4) N/C
The direction of the electric field E2 at the middle point is towards the negative charge.
Total electric field:
The total electric field is given by the addition of the electric field. The direction of the electric field is the same for both charges, hence total electric field E is,
E=E1+E2
E=4.32*10^(4) +4.32*10^(4)
E=8.64*10^(4) N/C
Learn more about electric field,
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