1.Four point charges Q1 - + 1 mC, Q2 = +1 mC and Q3 = -1 mC and Q4 = +1 mC are p

Question

1.Four point charges Q1 - + 1 mC, Q2 = +1 mC and Q3 = -1 mC and Q4 = +1 mC are placed in vertices of a square with 2 m long sides. Find electric field (magnitude and direction) and electric potential (magnitude and sign) in the center of the square. Find force acting upon the charge Q1. Find total potential energy of the interaction of the charges. 2. An electric field of 500 V/m is induced inside a two-plate capacitor with plate area of 0.5 m^2 by a 200 V voltage. Find capacitance of the capacitor and the charge on its plates. 3. Find currents (values and directions) in all resistors of the circuit given below. Find total electric power developing in this circuit. Neglect internal resistance of the batteries. 4. A voltmeter with internal resistance 30 Ohm and ammeter with internal resistance I Ohm are going to be use for the measurements of the voltage across the resistor R2 and the current in the resistor R3. Predict the reading of the devices. 5. How much time will it take to rich a I mA current in resistor RI in the circuit shown below

Explanation / Answer

Q1. let the square has following vertices:

Q1 at A(0,0)

Q2 at B(2,0)

Q3 at C(2,2)

Q4 at D(0,2)

first coordinate is x coordinate and second coordinate is y coordinate.

center of square is at (1,1)

part a.

field due to Q1:

as Q1 is positive, electric field due to Q1 will be directed away from Q1

and towards the center.

vector along the electric field=(1,1)-(0,0)=(1,1)

distance between the two points=sqrt(1^2+1^2)=1.4142 m

then unit vector along the electric field=(1,1)/1.4142=(0.7071,0.7071)

magnitude of electric field=9*10^9*Q1/distance^2

=9*10^9*10^(-3)/1.4142^2=4.5*10^6 N/C

in vector notiation, electric field due to Q1 at center of the square

is given by E1=4.5*10^6*(0.7071,0.7071) N/C

field due to Q2:

as Q2 is positive, electric field due to Q2 will be directed away from Q2

and towards the center.

vector along the electric field=(1,1)-(2,0)=(-1,1)

distance between the two points=sqrt(1^2+1^2)=1.4142 m

then unit vector along the electric field=(-1,1)/1.4142=(-0.7071,0.7071)

magnitude of electric field=9*10^9*Q2/distance^2

=9*10^9*10^(-3)/1.4142^2=4.5*10^6 N/C

in vector notiation, electric field due to Q2 at center of the square

is given by E2=4.5*10^6*(-0.7071,0.7071) N/C

field due to Q3:

as Q3 is negative, electric field due to Q3 will be directed towards Q3

and away from the center.

vector along the electric field=(2,2)-(1,1)=(1,1)

distance between the two points=sqrt(1^2+1^2)=1.4142 m

then unit vector along the electric field=(1,1)/1.4142=(0.7071,0.7071)

magnitude of electric field=9*10^9*Q3/distance^2

=9*10^9*10^(-3)/1.4142^2=4.5*10^6 N/C

in vector notiation, electric field due to Q3 at center of the square

is given by E3=4.5*10^6*(0.7071,0.7071) N/C

field due to Q4:

as Q4 is positive, electric field due to Q4 will be directed away from Q4

and towards the center.

vector along the electric field=(1,1)-(0,2)=(1,-1)

distance between the two points=sqrt(1^2+1^2)=1.4142 m

then unit vector along the electric field=(1,-1)/1.4142=(0.7071,-0.7071)

magnitude of electric field=9*10^9*Q4/distance^2

=9*10^9*10^(-3)/1.4142^2=4.5*10^6 N/C

in vector notiation, electric field due to Q4 at center of the square

is given by E4=4.5*10^6*(0.7071,-0.7071) N/C

hence total electric field at the center =E1+E2+E3+E4

=2*4.5*10^6*(0.7071,0.7071) N/C

magnitude =9*10^6 N/C

direction is 45 degree with +ve x axis.

part b:

electric potential due to a charge q at a distance d is given as k*q/d

as here all the charges are at same distance from the center.

net electric potential=9*10^9*(Q1+Q2+Q3+Q4)/d

where d=1.4142 m

electric potential=9*10^9*2*10^(-3)/1.4142=1.278*10^7 N/C

part C:

force on Q1 due to Q2:

as Q1 and Q2 are both positive, they will repel each other.

hence force on Q1 will be towards Q1 and away from Q2.

vector along the force=(0,0)-(2,0)=(-2,0)

distance=2 m

unit vector along the direction of force=(-2,0)/2=(-1,0)

force magnitude=9*10^9*Q1*Q2/distance^2

=9*10^9*10^(-3)*10^(-3)/4

=2250 N

hence in vector notation, force =F1=2250*(-1,0) N

force on Q1 due to Q3:

as Q1 and Q3 are of opposite sign , they will attract each other.

hence force on Q1 will be towards Q3 and away from Q1.

vector along the force=(2,2)-(0,0)=(2,2)

distance=sqrt(2^2+2^2)=2.8284 m

unit vector along the direction of force=(2,2)/2.8284=(0.7071,0.7071)

force magnitude=9*10^9*Q1*Q3/distance^2

=9*10^9*10^(-3)*10^(-3)/2.8284^2

=1125 N

hence in vector notation, force =F2=1125*(0.7071,0.7071) N

force on Q1 due to Q4:

as Q1 and Q4 are both positive, they will repel each other.

hence force on Q1 will be towards Q1 and away from Q4.

vector along the force=(0,0)-(0,2)=(0,-2)

distance=2 m

unit vector along the direction of force=(0,-2)/2=(0,-1)

force magnitude=9*10^9*Q1*Q4/distance^2

=9*10^9*10^(-3)*10^(-3)/4

=2250 N

hence in vector notation, force =F3=2250*(0,-1) N

hence net force=F1+F2+F3=2250*(-1,0)+1125*(0.7071,0.7071)+2250*(0,-1)

=1454.5*(-1,-1) N

magnitude =1454.5*sqrt(2)=2056.973 N

direction is 45 degree below -ve x axis

part d:

total potential energy of the interaction of the charges=(k*Q1*Q2/2)+(k*Q1*Q3/1.4142)+(k*Q1*Q4/2)+(k*Q2*Q3/2)+(k*Q2*Q4/1.4142)+(k*Q3*Q4/2)=0 J

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