Question

A regular hexagon of side 10 cm has a change 5 micro coulomb at each of its vertices. calculate the potential at the centre of the hexagon.​

Answer 1

Answer:

V = 2.7 × 10⁶ V

Step-by-step explanation:

Given:

• Side of hexagon = 10 cm = 0.1 m
• Charge at each vertices = 5 μC = 5 × 10⁻⁶ C

To Find:

• Potential at the centre of the hexagon

Solution:

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

$\boxed{\tt V=k\times \dfrac{Q}{r} }$

where K = 1/4πε₀ = 9 × 10⁹ Nm²C⁻² , Q = charge, r = distance

Now potential at the centre of the hexagon due to a single charge is given by,

$\tt V= \dfrac{9\times 10^9\times 5\times 10^{-6}}{0.1}$

$\tt V=\dfrac{45\times 10^3}{0.1}$

$\tt V = 450\times 10^3 \: V$

Now potential due to 6 charges is given by,

$\tt V= 6V$

$\implies \tt 6\times 450\times 10^3$

$\implies \tt 2700\times 10^3$

$\implies \tt 2.7\times 10^6 \: V$

Hence the electric potential is 2.7 × 10⁶ V.

Answer 2

Question:-

• A regular hexagon of side 10 cm has a change 5 micro coulomb at each of its vertices. calculate the potential at the centre of the hexagon.

To Find:-

• Calculate the potential at the centre of the hexagon.

Solution:-

Here ,

$\tt\implies \: V = K \times \dfrac { Q } { r }$

$\tt\implies \: V = \dfrac { 9 \times { 10 }^{ 9 } \times 5 \times { 10 }^{ -6 } } { 0.1 }$

$\tt\implies \: V = \dfrac { 45 \times { 10 }^{ 3 } } { 0.1 }$

$\tt\implies \: V = 450 \times { 10 }^{ 3 }$

Now ,

Potential Energy:-

$\tt\implies \: V = 6V$

$\tt\implies \: V = 6 \times 450 \times { 10 }^{ 3 }$

$\tt\implies \: V = 2.7 \times { 10 }^{ 6 }$

Hence ,

• Potential Energy is $\tt \: 2.7 \times { 10 }^{ 6 }$