1-)Two positive point charges q are placed on the x-axis, one at x=a and one at x=-a.
A) Find the electric field at x=0 .
B) Derive an expression for the electric field at points on the x-axis, where -a<x<a.
Express your answer in terms of the variables q,x ,a and appropriate constants. Ex=
C) Derive an expression for the electric field at points on the x-axis, where x>a.
Express your answer in terms of the variables q ,x , a and appropriate constants. Ex=
D) Derive an expression for the electric field at points on the x-axis, where x <-a.
Express your answer in terms of the variables q, x,a and appropriate constants. Ex=
2) Two particles with positive charges q1 = 0.550nC and q2= 8.20nC are separated by a distance of 1.20m .At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?
Anwer by stating L= …………………….. m from q1, between the charges
3) A point charge q1=-4.00nC is at the point x=0.60m, y=0.80m , and a second point charge q2=+6.00nC is at the point x=0.60m , y=0.
A) Calculate the magnitude of the net electric field at the origin due to these two point charges.
B) Calculate the direction of the net electric field at the origin due to these two point charges. (clockwise from +-x axis)
3 Answers
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1) Two positive point charges q are placed on the x-axis, one at X=a and one at X= – a.
a) At x=0, the electric field ( E ) is zero because electric field due to q at (a) is equal and opposite to the electric field due to q at (-a)
(b) the electric field E at points on the x-axis, where – a< x< a is resultant of electric field E1 due to q at (a) and electric field E2 due to q at (- a)
As E1 and E2 are in opposite direction,their resultant is equal to difference of their magnitudes
E = kq /(a-x )^2 – kq /(a+x )^2
E =kq [ (a+x )^2 – (a-x )^2] /(a-x )^2 *(a+x )^2
E =kq*4ax /(a^2 – x^2)^2
(c) The electric field at points on the x-axis, where x> a is resultant of electric field E1 due to q at (a) and electric field E2 due to q at (- a)
As E1 and E2 are in same direction,their resultant is equal to sum of their magnitudes
E = kq /(x +a )^2 + kq /(x – a )^2
E = kq [(x +a)^2 + (x- a)^2] /(x+a )^2*(x-a )^2
E = 2 kq [x^2 + a^2] /( x^2 – a^2 )^2 along positive direction of x axis
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d) The electric field at points on the x-axis, where x< – a is resultant of electric field E1 due to q at (a) and electric field E2 due to q at (- a)
As E1 and E2 are in same direction,their resultant is equal to sum of their magnitudes
E = kq / (a-x )^2 + kq /( a+x )^2
E = kq [(a+x )^2 + (a -x)^2] /(a-x )^2*(a+x )^2
E = 2 kq [a^2 + x^2] /( a^2 – x^2 )^2 along negative direction of x axis
Two particles with positive charges q1 = 0.550nC and q2= 8.20nC are separated by a distance of 1.20m .At point L meter from q1 along the line connecting the two charges the total electric field due to the two charges equal to zero
E due to q1 is equal in magnitude to E due to q2 but in opposite direction
As both q1 and q2 are positive, the field is zero at a point in between them
As distance from q1 is L , distance from q2 is (1.2 – L)
kq1/L^2 = kq2/(1.2 – L)^2
(1.2 – L) / L= sq rt (q2 / q1)
(1.2 – L) = L* sq rt (q2 / q1)
1.2 = L [1+sq rt (q2 / q1)]
L = 1.2 / [1+sq rt (q2 / q1)]
L = 1.2 / [1+sq rt (8.20 / 0.550)]
L = 1.2 / [1+sq rt ( 14.909 )]
L = 1.2 / [1+ 3.8612]
L = 1.2 / [ 4.8612 ]
L=0.2469 meter from 0.550 nC
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