Two positive point charges q are placed on the x -axis, one at x = a and one at

Question

Two positive point charges q are placed on the x-axis, one at x=a and one at x=?a.

1)Derive an expression for the electric field at points on the x-axis, where x>a.

2)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.

3)A small sphere with mass m carries a positive charge q and is attached to one end of a silk fiber of length L. The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density ?.

Assume that the sphere is in equilibrium and find the angle that fiber makes with the vertical sheet.

Explanation / Answer

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)
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(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
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(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

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