Finding the charge

To find the electric field caused by ring, it is only a one-step integration. We know that the electric field for a continuous distribution of charges is given by

$E=k \int \frac{dq}{x^{2}}$

where x is the distance between an infinitesimally small charge dq and the point whereby the electric field is measured.

Since we only need to consider the horizontal components of the electric field caused by every infinitesimally small charge dq as the vertical components are being cancelled out, the E field is therefore

$E=k\int \frac{dq \cos \theta }{x^{2}}$

$=k\int \frac{ldq }{ {(r^{2} + l^{2})}^{1.5}}$

$=k\frac{lQ }{ {(r^{2} + l^{2})}^{1.5}}$

Whereby l is the distance between the point and the centre of the ring and r is the radius of the ring, Q is the charge on the ring.

Now let q be the charge applied onto the 2nd ring. Since the net electric field is zero, we can equate the electric field caused by both rings

$\frac{k lQ }{ {(r^{2} + l^{2})}^{1.5}} =\frac{k (1-l)q }{ {(r^{2} + (1-l)^{2})}^{1.5} }$

Substituting Q= 1nC, l = 0.25 and r = 0.25 will give us q= 3.73nC .

Let's analyze the influence of one ring on the free charge:

Let $Q$ be the total charge on the ring; $q$ the free charge $x$ the distance ring's center-free charge; $R$ the ring radius; $λ$ the ring's linear density of charges; $Θ$ the angle formed by $x$ and the distance from a point on the ring and the free charge.

Considering an infinitesimally small point with charge $dQ$ , we can express $dQ$ as $dQ=dL λ$ . This point would generate a horizontal force dF given by(Coulomb's law):

$dF=\frac {k q}{x²+R²} dQ cosΘ \Rightarrow$ $dF=\frac {k q}{x²+R²} dL \lambda cos Θ$ (note that on the vertical axis, because of the ring's symmetric shape, the resultant force is 0)

Integrating for the whole ring:

$F=\int_0^{2 \pi R} \frac {k λ q}{x²+R²} dL cos Θ$

Solving:

$F=\frac {k λ q}{x²+R²} cos Θ \int_0^{2 \pi R} dL$

$F=\frac {k λ q}{x²+R²} cos Θ 2 \pi R$

As $cos Θ = \frac{x}{\sqrt{x²+ R²}}$ and $\lambda = \frac{Q}{2 \pi R}$

$F=\frac {k Q q}{x²+R²} \frac{x}{\sqrt{x²+ R²} (2 \pi R)} 2 \pi R \Rightarrow F=\frac {k Q q}{x²+R²} \frac{x}{\sqrt{x²+ R²} }$

F will be the force generated by one generic ring on the free charge. If we use the index "1" for the ring closer to the charge and "2" for the other one:

$F_1 = F_2$

$\frac {k Q_1 q}{x²_1+R²} \frac{x_1}{\sqrt{x²_1+ R²} } = \frac {k Q_2 q}{x²_2+R²} \frac{x_2}{\sqrt{x²_2+ R²} }$

$\frac { Q_1 }{x²_1+R²} \frac{x_1}{\sqrt{x²_1+ R²} } = \frac { Q_2 }{x²_2+R²} \frac{x_2}{\sqrt{x²_2+ R²} }$

Replacing the values given by the question:

$\frac { 1 }{25²+25²} \frac{25}{\sqrt{25²+ 25²} } = \frac { Q_2 }{75²+25²} \frac{75}{\sqrt{75²+ 25²} }$

$Q_2 = 3.726$

We can determine that the electric field intensity of a charged ring along its axis is equal to $\frac {kQx}{a^3}$ where k is Coulomb's constant, Q is the charge of the ring, x is the distance along the axis from the point charge to the center of the ring, and a is the distance from the point charge to a point on the ring. We can then plug in the values for the ring of known charge. By Pythagorean theorem, a is equal to $\sqrt{ .25^2+.25^2}$ = 0.3535m and $\frac {(8.99*10^9)(1*10^-9)(0.25)}{.3535^3} = 50.86$

Setting this value equal to the electric field of the second ring, we know that x = 0.75m and $a = \sqrt{.25^2+.75^2} = .7906m$ Plugging these values into the equation $\frac {kQx}{a^3}$ and solving for Q gives us a value of 3.727*10^-9 coulombs, or $3.727 nanocoulombs$

For every infinitesimal amout of charge in the ring with charge 1 nC, we can find an equivalent infinitesimal charge in the second ring which counters the force of the first ring's charge.

The eletric field on the ball generated by the infinitesimal amount of charge in the the rings is proportional to $1/r^2$ , $r$ is the distance from the ball to the infinitesimal charge.

Remember: by the symmetry, we must look after only for the field component in the axis of the rings. If $dE$ is the infinitesimal field generated, his component in the axis of the rings, in a distance $x$ from the center of the ring is related by: $dE \propto \frac{dq \times x }{(x^2+R^2)^{3/2}}$

If we integrate this for all charges in the ring, we have for some constant $k$ : $E \propto \frac{xkQ }{(x^2+R^2)^{3/2}}$

So in the equillibrium position, this field generated by both rings must be equal. Using the rings configuration, we have:

$\frac{Q_1 \times x }{(R_1^2+x^2)^{3/2}} = \frac{Q_2 \times (D-x) }{(R_2^2+(D-x)^2)^{3/2}}$

Now,

$R_1=R_2=25cm$

$x = 25cm$

$D = 1m$

$Q_1 = 1nC$

and from the equation above, we get $Q_2 = 3.73 nC$

The electric field of a charged ring of CHARGE Q, radius R, at a distance X from its centre on its axis is:- E=kQX/(R.R+X.X)^(3/2) ............Where k is a constant. k(1nC)(.25)/((.25)(.25)+(.25)(.25))^(3/2)=kQ(1-0.25)/((.25)(.25)+(.75)(.75))^(3/2) 3Q=(5)^(3/2) Q=3.73 nC

As we can see both the rings will exert an electric force in opposite directions on the particle.

Now, Electric field at a point at a distance x due to a ring of radius R can be given as E = $\frac {KQx} {(x^2+R^2)^\frac {3} {2}}$ where K = $\frac {1} {4\pi\epsilon_0}$

Therefore the force on the particle of charge q due to the known charge ring will be: F $_1$ = $\frac {KQ_1qx} {(x^2+R^2)^\frac {3} {2}}$ and force due to the unknown charge on the particle will be F $_2$ = $\frac {KQ_2q(1-x)} {((1-x)^2+R^2)^\frac {3} {2}}$

Now for equilibrium, $\frac {KQ_1qx} {(x^2+R^2)^\frac {3} {2}}$ = $\frac {KQ_2q(1-x)} {((1-x)^2+R^2)^\frac {3} {2}}$

Putting in the values we get,

$Q_2 = \frac {5\sqrt{5}} {3}$ nC

Solving this we get $Q_2 = 3.726$ nC

Suppose that $q_{2}$ is charge between $q_{1}=1 nC$ and $q_{3}=...?$ . Distance $q_{2}$ with $q_{1}$ is $25 cm$ , and Distance $q_{2}$ with $q_{3}$ is $75 cm$ and all radii $x_1=x_2$ If the condition equilibrium, We get: $E_{1,2}=E_{2=3,2}$ $\frac{k\cdot x_1\cdot q_1\cdot q_2}{\sqrt{(r_1^2+x_1^2)^{3}}}=\frac{k\cdot x_2\cdot q_3\cdot q_2}{\sqrt{(r_2^2+x_2^2)^{3}}}$ \frac{1nC\cdot 25 cm}{\sqrt{(25cm^2+25cm^2)^{\frac{3}{2}}}}=\frac{q_{3}\cdot 75 cm}{\sqrt{(25cm^2+75cm^2)^{\frac{3}{2}}}})\ And we get, \(q_{3}=\frac{5\sqrt{5}}{3}=3.73

First, we have to start off by drawing a simple diagram. We draw 2 discs, with their centers placed on the same axis. Let's say that right disc has the charge of 1 nC, and the right disc has the unknown charg we are asked to find. 25 cm from the rigt disc we place a test charge Qo. On our test charge we'll examine the forces acting on it from 2 discs. Now, the main part of this problem is to replace the whole charge of a disc with one point charge, with the same value. The best position to place the total charge of the disc is the highest point on the disc, so that the radius of the disc and an axis make an angle of 90 degrees. So after replacing we end up having a system of 3 charges: Q1 - whose walue is given, Q2 - whose value has to be found and Q0 - our test charge that will cancel out later in our equations.

The first step would be to calculate the distances from our test charge Qo and one of the charges, namely x1 and x2. Using Pythagorean theorem we have:

x1 = \sqrt{0.25^2 + 0.25^2}\ = 0.35 meters and x2 = \sqrt{0.25^2 + 0.75^2}\ = 0.79 meters.

Then, we take a look at the forces acting on the test charge from Q1 and Q2. We have: F1 = k Q1Qo/x1^2 and F2 = k Q2Qo/x2^2. But we cannot simply say that F1 = F2 and find out what's the charge Q2. Instead of that, we have to find the projection of 2 forces on the axis that connects the centers of 2 discs. We have:

F1 cos \alpha = F2 cos \beta, where cos alpha = 0.25/0.35, and cos beta = 0.75/0.79

Putting that in our previous equation we have:

(k Q1Qo/x1^2)* 0.25/0.35 = (k Q2Qo/x2^2) * 0.75/0.79

k's and Qo's cancel out and we are left with:

(Q1/x1^2) * 0.25/0.35 = (Q2/x2^2) * 0.75/0.79

from which we get that Q2 = 3.752 nC.

First we can calculate the electric field on the axis of a single ring charged with $Q$ . If we divide the ring into infinitesimal pieces so that each one of them can be approximated as a point charge $dQ$ , the field from it has the value $E = \frac{dQ}{4\pi\epsilon_0 d^2}$ . The overall field at the point $h$ on the ring’s axis would be obtained by summing up all the contributions. However, since the ring is symmetrical, the x components cancel out, and we can only add the y components. Since all the y components at a point h are at the same distance $d = \sqrt{h^2+r^2}$ , the sum gives us $E_h = \frac {Q h}{4 \pi \epsilon_0 {(r^2+h^2)}^{\frac{3}{2}}}$ . Since our charge was in equilibrium, $E_1(\frac{L}{4})= E_2(\frac{3L}{4})$ . Plugging in the numbers and solving for $Q_2$ we get $Q_2=3.73 Q_1 = 3.73~\mbox{nC}$ .