Question

A regular pentagon with side length 1 units is taken and then a pentagram is cut out from it .

A free point charge Q = 1 × 10⁶ C is placed directly above the centre of the pentagram at a distance of -

$\displaystyle \sf{ \dfrac{ 1}{ 2 } \sqrt[2]{ \dfrac{ 5 - \sqrt{5} }{ 10 } } } \: units .$ .


Calculate the total electric flux through the pentagram . ​

Answer 1

$\underline{\underline{\sf{\maltese\:\:Question}}}$

A regular pentagon with side length 1 units is taken and then a pentagram is cut out from it.  A free point charge Q = 1 × 10⁶ C is placed directly above the centre of the pentagram at a distance of

$\sf{\displaystyle\frac{1}{2} \sqrt[\sf{2}]{\frac{5-\sqrt{5}}{10}} \textsf{ units. }}\textsf{Calculate the total electric flux through the pentagram.}$

$\underline{\underline{\sf{\maltese\:\:Given}}}$

• A regular pentagon with side length 1 units is taken and then a pentagram is cut out from it
• The free point charge inside the pentagram is 1 × 10⁶ C

$\underline{\underline{\sf{\maltese\:\:To\:Find}}}$

• The total electric flux through the pentagram

$\underline{\underline{\sf{\maltese\:\:Intoduction}}}$

• Electric flux is the rate of flow of the electric field through a given area. The total number of electric field lines passing a given area in a unit time is defined as the electric flux.  Electric flux is proportional to the number of electric field lines going through a virtual surface. Electric flux is a scalar quantity. It is a scalar because it is the dot product of two vector quantities, electric field and the perpendicular differential area. SI unit of Electric flux Nm²/C or Nm²C⁻¹

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

• The total electric flux through the pentagram = 7.5 × 10¹⁶ Nm²/C

$\underline{\underline{\sf{\maltese\:\:Calculation}}}$

Step 1) Write the expression of the electric flux.

$\textsf{Expression of electric flux : }\sf{\phi=\dfrac{k Q}{r^{2} A}}$

• Where, k is the Coulomb's constant, r is the distance and A is the area of pentagram.

Step 2) Substitute the values in the above expression.

Substitute 9 × 10⁹ Nm²/C² for k, 1 × 10⁶ C for Q, 1.72 for A and

$\sf{\displaystyle\frac{1}{2} \sqrt[\sf{2}]{\frac{5-\sqrt{5}}{10}}\textsf{ for r}}$

$\sf{\phi=\dfrac{k Q}{r^{2} A}}$

$:\implies\sf{\displaystyle\phi=\dfrac{(9\times 10^9\:Nm^2/C^2)(1\times 10^6 \:C) }{\left(\frac{1}{2} \sqrt[2]{\frac{5-\sqrt{5}}{10}}\right)^{2} (1.72)}}$

Step 3) Simplify the Expression

$:\implies\sf{\displaystyle\phi=\dfrac{(9\times 10^9\:Nm^2/C^2)(1\times 10^6 \:C) }{\left(\frac{1}{2} \sqrt[2]{\frac{5-\sqrt{5}}{10}}\right)^{2} (1.72)}}$

$:\implies\sf{\displaystyle\phi=\dfrac{(9\times 10^9\:Nm^2/C^2)(10^6 \:C) }{\left(\frac{1}{4}\times \frac{5-\sqrt{5}}{10}\right) (\frac{172}{100})}}$

$:\implies\sf{\displaystyle\phi=\dfrac{9\times 10^9\:Nm^2/C\times 10^6}{\left(\frac{1}{4}\times \frac{5-\sqrt{5}}{10}\right) (\frac{172}{100})}}$

$:\implies\sf{\displaystyle\phi=\dfrac{9\times 10^{9+6}}{\left(\frac{1}{4}\times \frac{5-\sqrt{5}}{10}\right) (\frac{172}{100})}}\sf{Nm^2/C}$

$:\implies\sf{\displaystyle\phi=\dfrac{9\times 10^{15}}{\left(\frac{1}{4}\times \frac{5-\sqrt{5}}{10}\right) (\frac{172}{100})}}\sf{Nm^2/C}$

$:\implies\sf{\displaystyle\phi=\frac{9\times 10^{15}}{\frac{1\times (5-\sqrt{5})\times 172}{4\times 10\times 100}}\:\sf{Nm^2/C}}$

$:\implies\sf{\displaystyle\phi=\frac{9\times 10^{15}}{\frac{(5-\sqrt{5})\times 172}{4000}}\:\sf{Nm^2/C}}$

$:\implies\sf{\displaystyle\phi=\frac{9\times 10^{15}}{(5-2.236)\times 0.43}}\:\sf{Nm^2/C}$

$:\implies\sf{\displaystyle\phi=\frac{9\times 10^{15}}{2.77\times 0.43}}\:\sf{Nm^2/C}$

$\boxed{\boxed{:\implies\sf{\displaystyle\phi=75\times 10^{15}}\:\sf{Nm^2/C}}}$

Therefore,

• The total electric flux through the pentagram = 7.5 × 10¹⁶ Nm²/C

Answer 2

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