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arguments and heated arguments have a lot of difference my friend.

I think you mean that the charge density is proportional to the radius. So assuming the above, we have

$\rho$ = kR.

Assuming its the space charge distribution, the charge dq in a spherical shell of thickness dr is $dq = 4\pi kR^3 dR$

Hence Q = $\int^R_0 4\pi kR^3\, dR$ .

Q = $\pi kR^4$

Hence, using, $V_{center} = \frac{Q}{4\pi \epsilon_o R}$ , you can get your answer

The answer has been given by allamraju, but, in case you are interested in knowing how that formula is obtained, here goes:

The satellite as such moves around the earth in an elliptical orbit, which is assumed to be almost circular for simplicity. Now, any object moving in a circle requires that a force acts on it towards the center which keeps it moving in that circle. (For example, when you whirl a stone tied to a string, it moves in a circle because the string keeps the stone from breaking off and moving away. If the string ruptures, the stone would move away, wouldn't it?)

This central force is called the centripetal force, and its magnitude is given by $\frac{mv^2}{R}$ , where R is the radius of the orbit, and v is the velocity of the object, Here, for the satellite, the force acting towards the center of the orbit is nothing but the gravitational force of attraction on the satellite due to the earth, and we know that its magnitude is given by $\frac{GMm}{R^2}$ .

Hence, for the satellite to move in a circular path, we must have

$\frac{GMm}{R^2} = \frac{mv^2}{R}$

Which boils down to the result $v = \sqrt{\frac{GM}{R}}$ .

If you've understood this, even if you forget the formula, you'll be able to derive it in seconds.

Differentiate it and equate it to zero!

Please post the question clearly buddy

No buddy, Potential at the center = Potential at the surface of the sphere.

Chill guys! Its not very nice to see two of goIIT's best brains arguing this way...

My last reply was distorted. Here goes :

we have $tan\theta = c$ and $\frac{a}{2u^2 cos^2 \theta} = b$

Now you can solve for u.

The given charge density is $\rho$ = A(R-r)

Now consider a shell sandwiched between two spheres of radii r and r+dr. Its volume = $4\pi r^2 dr$ .

So, the charge enclosed in this shell is $dq = 4\pi r^2 A(R-r)dr$ .

The total charge Q is given by q = $\int^R_0 4\pi r^2 A(R-r)\, dr$ .

Using the above, you can find A.

To find the electric field at the desired points, simply use Gauss's law. It is fairly simple, just basic integration. For calculating the field inside a point on the sphere, you will have to calculate the charge enclosed by the part of the sphere enclosed.

The given equation is analogous to that of a projectile fired from the ground. Compare the given equation with

Post the questions please

Use mesh analysis.

-2i + 1 + 1 - 2i + 1 - 2i = 0

3 = 6i

i = 0.5A

The formula for capillary rise is actually h = 2Scos@/rdg

You can conserve angular momentum about the point of contact of the two discs. Alternatively, you can solve it using conventional rotational mechanics.

akku's solution is correct, aka's solution is wrong.