solution:
E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})
=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})
x=\frac{r}{a}
infinite case,
Ei=\frac{r}{\epsilon0}
\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})
we have to find x when,
ei-e\delta =1% ,y=ei=1/100 ei
or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei
\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}
4x^2+1 =10^4
x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50
\therefore \frac{r}{a}\approx 50
Answer:
(a) The force between them quadruples
Explanation:
According to coulomb's law, initial force between the two charged objects is given as;

where;
k is coulomb's constant
q₁ is the charge on the first object
q₂ is the charge on the second object
r is the distance between the two objects
When the charges on both objects are doubled, then;
q₁ = 2q₁
q₂ = 2q₂
Force between the two charged objects will become

Therefore, the force between them quadruples
Say we have a cylinder
that has a height of dx, we see that the cylinder has a volume of: <span>
<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx
Then, the weight of oil in this cylinder is:
Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.
Then, since the oil x feet from the top of the tank needs to
travel x feet to get the top, we have:
Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.
<span>Integrating from x1 to x2 ft gives the total work to be: (x1
= distance from top liquid level to ground level; x2 = distance from bottom
liquid level to ground level)</span>
<span>W = ∫ 1250π x dx
<span>W = 1250π ∫ x dx
W = 625π * (x2 – x1)</span></span></span></span>
<span>x2 = 14 ft + 15 ft = 29 ft</span>
x1 = 14 ft + 1 ft = 15
ft
<span>
W = 625π * (29^2 - 15^2)
<span>W = 385,000π ft-lbs
= 1,209,513.17 ft-lbs</span></span>
This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.

Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:

Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.

This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was

It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:

Solving for h gives us:

It doesn't depend on mass!
What are the following statements? If there's one that mention a description of current action, or motion, that's your answer.