please answer question 11 and 12 Three charges are arranged as shown in the figu

Question

please answer question 11 and 12

Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 5.28 nC at the origin. (Let r_12 = 0.290 m.) magnitude direction degree counterclockwise from the +x-axis J (a) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) (b) If a charge of -7.01 mu C is placed at this point, what are the magnitude and direction of the force on it? magnitude

Explanation / Answer

Q11.

force on q due to 6 nC :

as both charges are positive, force is repulsive in nature.

so force is directed towards -ve x axis.

force magnitude=k*q*(6 nC)/r12^2

where k=coloumb constant=9*10^9

force magnitude=9*10^9*5.28*10^(-9)*6*10^(-9)/0.29^2=3.3902*10^(-6) N

force on q due to -3 nC:

as both charges have opposite signs, force is attractive and hence towards negative y axis.

force magnitude=9*10^9*5.28*10^(-9)*3*10^(-9)/0.1^2=1.4256*10^(-5) N

total force in vector notation=(-3.3902 i -14.256 j)*10^(-6) N

force magnitude=sqrt(3.3902^2+14.256^2)*10^(-6)=14.654*10^(-6) N

force is in 3rd quadrant.

angle with -ve x axis=arctan(14.256/3.3902)=76.623 degrees

so counterclockwise angle with +ve x axis=180+76.623=256.62 degrees

Q12.

electric field at a distance of d from a charge of q =k*q/d^2

so total field due to the three charges

=(9*10^9*6*10^(-6)/0.02^2)-(9*10^9*1.5*10^(-6)/0.01^2)+(9*10^9*2*10^(-6)/0.03^2)

=2*10^7 N/C

part b:

force magnitude=charge*electric field

=7.01*10^(-6)*2*10^7=140.2 N

direction will be towards left.

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