Question

Electric charges 1000 micro coulomb are placed at point A and B respectively at a distance of 2 m from each other calculate the electric field at mid point of the line AB and at a point at equal distances of 4 m from each charger

Answer 1

Answer:

Electric charges -+1000 microcoulomb are placed at points A and B respectively at a distance of 2m from each other.Calculate the electric feilds at (a)midpoint of the line AB and (b)at a point at equal distances of 4m from each charge(ans (a)1.8*10 N/C along AB ..(b)2.8 *10 to the power 5 N/C,parallel to AB)

Since electric field is a vector quantity, we will not consider the signs of the charges in this question.

We know that electric field is given as:

E = kq/r2

Here k is the constant = 9*109

q is the charge = 1000 *10-6C

r is the distance of the charge from mid point = 1m

let electric field due to 1 charge be E1 and due to the other be E2­. Since both the charges are equal in magnitude, E1 = E2. Also, the final E will be E1 + E2

hence 2kq/r2,

2*9*109 *103 *10-6

1.8 *10-5N/C

2nd case

Here the point at which the electric field has to be found forms an isosceles triangle with both the charges each 4m from the equatorial point .

This point lies on the equatorial point.

E = k*2aq/(r2 +a2)3

Here r = altitude = √15

a = 1m = ½ the distance from each charge.

K = 9*109

2a = 2m

Substituting the values in the above formula we get,

E = 2.8 *105N/C.

Explanation:

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Answer 2

Answer:1.8*10^7N/C

2.8*10^5N/C

Explanation:

Electric charges 1000 micro coulomb are placed at point A and B respectively at a distance of 2 m from each other calculate the electric field at mid point of the line AB and at a point at equal distances of 4 m from each charger

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mananmadani53

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Answer:

Electric charges -+1000 microcoulomb are placed at points A and B respectively at a distance of 2m from each other.Calculate the electric feilds at (a)midpoint of the line AB and (b)at a point at equal distances of 4m from each charge(ans (a)1.8*10 N/C along AB ..(b)2.8 *10 to the power 5 N/C,parallel to AB)

Since electric field is a vector quantity, we will not consider the signs of the charges in this question.

We know that electric field is given as:

E = kq/r2

Here k is the constant = 9*109

q is the charge = 1000 *10-6C

r is the distance of the charge from mid point = 1m

let electric field due to 1 charge be E1 and due to the other be E2­. Since both the charges are equal in magnitude, E1 = E2. Also, the final E will be E1 + E2

hence 2kq/r2,

2*9*109 *103 *10-6

1.8 *10-5N/C

2nd case

Here the point at which the electric field has to be found forms an isosceles triangle with both the charges each 4m from the equatorial point .

This point lies on the equatorial point.

E = k*2aq/(r2 +a2)3

Here r = altitude = √15

a = 1m = ½ the distance from each charge.

K = 9*109

2a = 2m

Substituting the values in the above formula we get,

E = 2.8 *105N/C.