Question

In the figure, four charges, given in multiples of 4.00

In the figure, four charges, given in multiples of 4.00

Homework Answers

Answer #1

electric field is a vector quantity so we have to find both the magnitude and direction.

basic formula is kQ/r^2 where k = 9*10^9 and r = distance and Q =charge.

consider the two charges +5q and -5q

since the direction of electric field from a positive charge is from it to the point and that of a negative charge is from the point to itself. here the directions both add up

so calculate the electric field because of a 5q charge and double the magnitude and the direction will be 45 degrees to +ve x-axis in anticlock direction.

E1= 2(k5q/r^2)

here the distance from the 5q charge to centre is sqrt(2d^2)

therefore E1=2(k5q/2d^2) = 5kq/d^2.

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now there are two 3q charges unlike 5q charges they don't add up because they are of same sign or both are positively charged.

so the electric fields due to individual charges cancel out at the middle ie., at the centre therefore electric field due to 3q charges is zero

____________________________________________________________________________________________

there are 2 q charges on the y axis

as discussed above they two are of same sign and hence they cancel out their electric fields and the net due to both equals zero

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after that there is one +q charge on -ve x-axis and -2q charge on +ve x-axis so they two add up and the direction of the electric will be in +ve x-axis direction.

E2=kq/r^2 + k2q/r^2

here r is d

therefore E2= kq/d^2 +2kq/d^2 = 3kq/d^2 in +ve x-axis direction

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now as a result of all the charges we got two electric fieds

they are E1 with a direction of 45 degrees with +ve x-axis direction anticlockwise

and E2 along +ve x-axis direction.

E1= 5kq/d^2 = 5*9*10^9*4*10^-6/(1*10^-2)^2 = 180*10^7 N/C

the direction will be 45 degrees

so +ve x-axis component is E1cos45=127.28 *10^7 N/C

and +ve y-axis component is E1sin45=127.28*10^7 N/C

E2= 3kq/d^2 = 108*10^7 N/C along +ve x-axis

now net Ex component is (127.28+108)*10^7 N/C = 2.35*10^9 N/C

net Ey component is 1.27*10^9 N/C

net magnitude is sqrt(Ex^2 + Ey^2) = 2.67*10^9 N/C

this has a direction of arctan(1.27/2.35) =28.39 degrees with +ve x-axis in anticlock direction.

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