Question

The electric potential inside a charged spherical conductor of radius R is is given by V = keQ/R, and the potential outside is given by V = keQ/r. Using Er = -dV/dr, derive the electric field inside and outside this charge distribution. (Use the following as necessary: ke, Q, r, and R.)

Answer 1

Answer:

For outer points of shell

$E = \frac{k_eQ}{r^2}$

Now for inner point of shell

$E = 0$

Explanation:

As we know that out side the shell electric potential is given as

$V = \frac{K_e Q}{r}$

inside the shell the electric potential is given as

$V = \frac{K_e Q}{R}$

now we know the relation between electric potential and electric field as

$E = - \frac{dV}{dr}$

so we can say for outer points of the shell

$E = -\frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{r})$

$E = \frac{k_eQ}{r^2}$

Now for inner point again we can use the same

$E = - \frac{dV}{dr}$

$E = - \frac{d}{dr}(\frac{K_eQ}{R})$

$E = 0$

Answer 2

Electric field inside and outside of charged spherical conductor is electric force per unit charge.

• a) The electric field inside charge distribution is,
•    $E=-k_e\dfrac{Q}{r^2}$
• b) The electric field outside charge distribution is 0.

What is electric field?

The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.

Given information-

The radius of the spherical conductor is R.

The electric potential inside a charged spherical conductor is given by,

$V = k_e\dfrac{Q}{R}$

The electric potential outside a charged spherical conductor is given by,

$V = k_e\dfrac{Q}{r}$

As the relation given between the electric field and electric potential is,

$E_r=-\dfrac{dv}{dr}$           .....1

Let the above equation is equation one,

• a) The electric field inside charge distribution-

The electric potential inside a charged spherical conductor is given by,

$V = k_e\dfrac{Q}{r}$

Put this value of electric potential in the equation one as,

$E_r=-\dfrac{d}{dr}(k_e\dfrac{Q}{r})\\E=-k_e\dfrac{Q}{r^2}$

Hence, the electric field inside charge distribution is,

$E=-k_e\dfrac{Q}{r^2}$

• b) The electric field outside this charge distribution.

The electric potential outside a charged spherical conductor is given by,

$V = k_e\dfrac{Q}{R}$

Put this value of electric potential in the equation one as,

$E_r=-\dfrac{d}{dr}(k_e\dfrac{Q}{R})\\E=0$

Hence, the electric field outside charge distribution is 0.

• a) The electric field inside charge distribution is,
•    $E=-k_e\dfrac{Q}{r^2}$
• b) The electric field outside charge distribution is 0.

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