In: Physics
EXAMPLE 15.5 Electric Field Due to Two Point Charges
GOAL Use the superposition principle to calculate the electric field due to two point charges
PROBLEM Charge q1 = 7.00 μc is at the origin, and charge q2 = -5.00 μc is on the x-axis, 0.300 m from the origin.
(a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m. (b) Find the force on a charge of 2.00 x 10-8 C placed at P.
STRATEGY Follow the problem-solving strategy, finding the electric field at point P due to each individual charge in terms of x- and y-components, then adding the components of each type to get the x- and y-components of the resultant electric field at P. The magnitude of the force in part (b) can be found by simply multiplying the magnitude of the electric field by the charge
(a) Place a charge of -4.10 µC at point P and find the magnitude and direction of the electric field at the location ofq2 due to q1 = 7.40 µC and the charge at P.
magnitude | _______ N/C |
direction | _______ ° counterclockwise from the +x-axis |
(b) Find the magnitude and direction of the force on q2.
magnitude | _______ N |
direction | _______ ° counterclockwise from the +x-axis |