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**Hello. I have a problem calculating the electric field from spherical charge distribution. The exercise is:**

1. Homework Statement

1. Homework Statement

## Homework Equations

To solve the problem for $$ 0\le R < a$$ i tried 2 ways:

$$

\vec{E} = \frac{\vec{a_R}}{4\pi\epsilon_0}\int_v\frac{1}

{R^2}\rho dv

$$

and the second way

$$

V = \frac{1}{4\pi\epsilon_0}\int_v\frac{1}{R}\rho dv

$$

and

$$

\vec{E} = -\vec{\nabla}V

$$

3. The Attempt at a Solution

3. The Attempt at a Solution

Unfortunately, neither of the attempts worked.

Using first method, substituting the charge density from the text and using spherical coordinates to calculate the integral i got

$$

\vec{E} = \frac{Q\vec{a_R}}{4\pi\epsilon_0a^4}\int_v\frac{1}

{R}R^2\sin{\theta}dRd\theta d\phi = \frac{QR^2}{2\pi\epsilon_0a^4}\vec{a_R}

$$

Which is wrong. The correct answer is very close though, where there is a 4 instead of 2 in the answer above. No matter how much i looked i couldn't find where i missed the constant.

Using the second method, i got

$$

V = \frac{Q}{4\pi^2\epsilon_0a^4}\int_vR^2\sin{\theta}dRd\theta d\phi = \frac{QR^3}{3\pi\epsilon_0a^4}

$$

Which gives the electric field

$$

\vec{E} = -\vec{a_R} \frac{\partial{V}}{\partial{R}} = -\frac{QR^2}{\pi\epsilon_0a^4}\vec{a_R}

$$

The above answer is even more wrong than the previous one.

I would really appreciate if someone could tell me what i have done wrong.