Question

A ring has radius 30 cm and a charge of 2 nC. The electric
potential on the axis of the ring at a distance of 40 cm from
its centre will be​

Answer 1

The value of the electric potential is 36 V.

Option (3) is correct.

Explanation:

We are given that:

• The radius of the ring = 30 cm
• Charge on the ring = 2 nC
• Distance from ten center = 40 cm
• To Find: The Electric potential on the axis of the ring =?

Solution:

Vp = K q r / Sp

Vp = 9 x 10^9  x 2 x 10^-9 / 50 x 10^2 x 100^2

Vp = 18 / 50 x 10^-2 x 100^2  

Vp = 0.36 x 100  

Vp = 36 V

Thus the value of electric potential is 36 V.

Answer 2

Answer:

The correct answer is option (c) 36 V.

Given:

Radius of the ring, r = 30 cm = 30 × 10⁻² m = 0.3 m

Charge on the ring, q = 2 nC = 2 × 10⁻⁹ C

Distance from the centre, s = 40 cm = 40 × 10⁻² m = 0.4 m

Find:

The electric potential on the axis of the ring.

Solution:

The electric potential due to a charge at a point is given by,

$V = \frac{1}{4\pi \epsilon_o} \frac{q}{R}$

where V = Electric potential

$1/4\pi \epsilon_o$ = Constant = 9 × 10⁹ Nm²/C²

q = Charge

R = distance between charge and point.

Here, q = 2 × 10⁻⁹ C

R = $\sqrt{r^2+ s^2}$ = $\sqrt{0.3^2 + 0.4^2} = \sqrt{0.09 + 0.16} = \sqrt{0.25} = 0.5 m$

Putting all the values in the above formula, we get

$V = \frac{1}{4\pi \epsilon_o} \frac{q}{R}$

$V = (9\times 10^{9})\frac{2 \times 10^{-9}}{0.5} \\\\V = \frac{9 \times 2}{0.5} \\\\V = 18/0.5\\\\V = 36 V$

Hence, the electric potential on the axis of the ring at a distance of 40 cm from its center will be​ 36 volts.

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