2.13 A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

2.13

Length of the side of a cube = b

Charge at each vertices= q

d = diagonal of each face, d = $\sqrt{b^2+b^2}$ = b $\surd 2$

l = length of the solid diagonal, l = $\sqrt{d^2+b^2}$ = b $\surd 3$

If r is the distance from the centre of the cube to its corner, then r = l/2 = $\frac{\surd 3}{2}b$

The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.

V = $\frac{8q}{4\pi {\epsilon }_{0}r}$ = $\frac{8q}{4\pi {\epsilon }_0\frac{\surd 3}{2}b}$ = $\frac{4q}{\surd 3\pi {\epsilon }_{0}b}$

Therefore, the potential at the centre of the cube is $\frac{4q}{\surd 3\pi {\epsilon }_{0}b}$

The electric field at the centre of the cube, due to eight charges, gets cancelled, since the charges are distributed symmetrically. The electric field is zero at the centre.

<p><strong>2.13</strong></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1728972192phpQQBZHz_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1728972192phpQQBZHz.jpeg" alt="" width="152" height="155"></picture></div><div><p>Length of the side of a cube = b</p><p>Charge at each vertices= q</p><p>d = diagonal of each face, d =&nbsp;<span title="Click to copy mathml"><math><msqrt><mrow><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></msqrt></math></span>&nbsp;= b&nbsp;<span title="Click to copy mathml"><math><mo>√</mo><mn>2</mn></math></span></p><p>l = length of the solid diagonal, l =&nbsp;<span title="Click to copy mathml"><math><msqrt><mrow><msup><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>+</mo><msup><mrow><mrow><mi>b</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></mrow></msqrt></math></span>&nbsp;= b&nbsp;<span title="Click to copy mathml"><math><mo>√</mo><mn>3</mn></math></span></p><p>If r is the distance from the centre of the cube to its corner, then r = l/2 =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mo>√</mo><mn>3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mi>b</mi></math></span></p><p>The electric potential (V) at the centre of the cube is due to the presence of eight charges at the vertices.</p><p>V =&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>8</mn><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>r</mi></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>8</mn><mi>q</mi></mrow></mrow><mrow><mrow><mn>4</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mfrac><mrow><mrow><mo>√</mo><mn>3</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac><mi>b</mi></mrow></mrow></mfrac></math></span>&nbsp;=&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>4</mn><mi>q</mi></mrow></mrow><mrow><mrow><mo>√</mo><mn>3</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>b</mi></mrow></mrow></mfrac></math></span></p><p>Therefore, the potential at the centre of the cube is&nbsp;<span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>4</mn><mi>q</mi></mrow></mrow><mrow><mrow><mo>√</mo><mn>3</mn><mi>π</mi><msub><mrow><mrow><mi>ε</mi></mrow></mrow><mrow><mrow><mn>0</mn></mrow></mrow></msub><mi>b</mi></mrow></mrow></mfrac></math></span></p><p>The electric field at the centre of the cube, due to eight charges, gets cancelled, since the charges are distributed symmetrically. The electric field is zero at the centre.</p></div></div>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment