IIT JEE , AIEEE Forum - Physics , Electricity

monorahi_iit

Point charges +q and -q are located at the vertices of a square with diagonals 2l. Find the magnitude of the electric field strength at a point located symmetrically w.r.t to the vertex of the square at a distance x from the center.

prahalad

Since the point is located at a distance x from the center of the sqr (symmetrical to the vertex)

So the point is at a distance of sqrt(x^2+l^2) from the four corners.

There fore the magnitude of electric field due to -q and +q are

I E(-q) l = Kq/(x^2+l^2)*cos45 and l E(q) l =Kq/(x^2+l^2)*cos45

Therefore net field be=sqrt2Kq/(x^2+l^2)

If it help it ...any doubt ask me....

monorahi_iit

If somehow i could have acess to the diagram, i would be grateful. I am unable to place the poinmt and comprehend.

elessar_iitkgp

I am assuming that two positive charges are placed at two successive vertices and then two negative charges on the next two vertices. The point where the electric field is needed is at a distance x above the center of the square.

If you draw the force vectors on the unit charge, it can be seen that the components of the forces perpendicular to the plane of the square cancel each other. The net force is then along the the plane of the square. Lets prove this:

In these types of problems, its best to use the vector equation approach.
We will use the following:
E_Pq = (q(r_P - r_q))/4

₀|(r_P - r_q)|³where E_Pq is the electric field at P due to q.

First mark the charges on the square as 1(q),2(q),3(-q),4(-q) and the point as P.
Now, assume the point of intersection of the diagonals as the origin, the diagonal through charge 1 to be the X axis and the diagonal through charge 2 to be the Y axis. Let the Z axis be from the center of the square, towards the point P.
Net force on the unit charge,
E_P = E_P1+E_P2+E_P3+E_P4E_P1= (q(xk - l i ))/4

₀(l² + x²)^3/2E_P2= (q(xk - l j ))/4

₀(l² + x²)^3/2E_P3= (-q(xk + l i ))/4

₀(l² + x²)^3/2E_P4= (-q(xk + l j ))/4

₀(l² + x²)^3/2
Adding these up,
E_P = [q/(4

₀(l² + x²)^3/2)] x [xk - l i + xk - l j - xk - l i - xk - l j]
F₅ = - [2ql/(4

₀(l² + x²)^3/2)]x[i + j]
|F₅| = 2

2ql/(4

₀(l² + x²)^3/2)
|F₅| = ql/(

2

₀(l² + x²)^3/2) =

2ql/(2

₀(l² + x²)^3/2)
As we had suspected, the net force is directed along the plane of the square.

However if the charges are placed alternatel positive and negative,
1(q),2(-q),3(q),4(-q),
Then the net force seems to be zero as the forces seem to cancel out. Lets prove it analytically.
E_P1= (q(xk - l i ))/4

₀(l² + x²)^3/2E_P2= (-q(xk - l j ))/4

₀(l² + x²)^3/2E_P3= (q(xk + l i ))/4

₀(l² + x²)^3/2E_P4= (-q(xk + l j ))/4

₀(l² + x²)^3/2Adding these up,
E_P = [q/(4

₀(l² + x²)^3/2)] x [xk - l i - xk + l j + xk + l i - xk - l j]
E_P = 0
which is what we suspected.
Cheers

monorahi_iit

sir, but the answer is

2/2piepsilon zero(ql(l²+x²)^3/2

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