A positive charge of magnitude Q_1=6.5 nC is located at the origin. A negative c

Question

A positive charge of magnitude Q_1=6.5 nC is located at the origin. A negative charge Q_2=nC is located on the positive x-axis at x=3.5 cm from the origin. The point P is located y=19 cm above charge Q_2. Calculate the x-component of the electric field at point P due to charge Q_1. Write your answer in units of N/C. E_x, 1=_____ Calculate the y-component of the electric field at point P due to charge Q_1. Write your answer in units of N/C. Calculate the y-component of the electric field at point P due to the Charge Q_2. Write your answer in units of N/C. Calculate the y-component of the electric field at point P due to both charges. Write your answer in units of N/C. Calculate the magnitude of the electric field at point P due to both charges. Write your answer in units of N/C. Calculate the angle in degrees of the electric field at point P relative to the positive x-axis.

Explanation / Answer

Given that

charge Q1=6.5 nC

charge q2=-7.5 nC

distance x=3.5 cm

distance y=19.5 cm

basing on the concept of intensity of electric field

now we find the electric field differenct position at point P

the electric field at charge q1 to point P=>?E1=9*10^9*6.5*10^-9(0.035i+0.195 j)/(0.035^2+0.195^2)^3/2

=172.14(0.035 i+0.195 j)

=6.025 i+33.6 j

the electric field Ex1=6.025 N/c

the electric field Ey1=33.6 N/c

now we find the electric field at charge Q2 to point P

the electric field E2=Ey2=9*10^9*-7.5*10^-9/0.195^2

=-1775.15 N/c

now we find the net electric field

the net electric field Enet=(6.025i+33.6 j)-1775.15 j

=6.025 i-1741.55 j

x-commponent electric field of both charges Ex=6.025 N/c

y-component electric field of both charges Ey=-1741.55 N/c

magnetitude =1741.56 N/c

direction =tan^-1[-1741.55/6.025]=90.2 degree

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