In a coordinate system a charge 25×10-9C is placed at the origin of the coordinate, and a charge -25×10-9C at a point x=6m,y=0.what is the electric field and its direction at a point x= 3m, y= 4m?

Question

Answer

E at 3,4 will be resultant vector of the E vectors whose magnitude is kQ/25
Q=given charge
the two vectors will make an angle of 74 degree with each other
so the resultant direction will be parallel to y axis
still any doubt
..reply...regards

Answer

Calculation of electric field at point (3,4)

Step 1: Determine the distance between the charges

The distance between the charges can be calculated using the Pythagorean theorem as follows:

d = sqrt((6-0)^2 + (0-0)^2) = 6m

Step 2: Determine the electric field due to each charge

The electric field at point (3,4) due to the charge at the origin can be calculated using Coulomb's law as follows:

E1 = k*q1/r1^2 = (9*10^9 N*m^2/C^2)*(25*10^-9 C)/(5m)^2 = 45 N/C

The electric field at point (3,4) due to the charge at (6,0) can also be calculated using Coulomb's law as follows:

E2 = k*q2/r2^2 = (9*10^9 N*m^2/C^2)*(-25*10^-9 C)/(5m)^2 = -45 N/C

Step 3: Calculate the net electric field at point (3,4)

The net electric field at point (3,4) is the vector sum of the electric fields due to each charge. Since the charges are of equal magnitude, their electric fields have equal magnitudes but opposite directions, so the net electric field is:

E = E1 + E2 = 45 N/C - 45 N/C = 0 N/C

Since the net electric field is zero, the direction of the electric field at point (3,4) is undefined.

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