Question

Five charges, \(q\) each are placed at the corners of a regular pentagon of side 'a' (Fig.).



(i) What will be the electric field at \(O\), the centre of the pentagon?

(ii) What will be the electric field at \(O\) if the charge from one of the corners (say A) is removed?

(iii) What will be the electric field at \(O\) if the charge \(q\) at 𝐴 is replaced by \(-q\) ?

(iv) What will be the net electric field at centre, if pentagon is replaced by n-sided regular polygon with charge \(q\) at each of its corners?

Answer 1

(i) Step 1: Draw the diagram showing the directions of electric fields.



Step 2: Find the electric field at centre O.

Given, Sides of pentagon,
AB = BC = CD = DE = EA = a
Distance of sides from the center,
OA = OB = OC = OD = OE = r



As given in the figure, the magnitude of electric fields will be given by,

\( E_{A}~=~E_{B}~=~E_{C}~=~E_{D}~=~E_{E}~=~E~=~ \dfrac{kg}{r^{2}}\)

Therefore, the net Electric Field at point at O in x direction will be,
\( E_{x~net}~=~E_{E}\cos 18^{\circ}~+~E_{D}\cos 54^{\circ}~-~E_{B}\cos 18^{\circ}-~E_{C}\cos 54^{\circ}~\)

\( E_{x~net}~=~E~\cos 18^{\circ}~+~E~\cos 54^{\circ}~-~E~\cos 18^{\circ}-~E~\cos 54^{\circ}~\)

\( E_{x~net}~=~0 \)

Now, the net Electric Field at point O in y direction will be,

\( E_{y~net}~=~E_{C}\cos 36^{\circ}~-~E_{D}\cos 36^{\circ}~-~E_{E}\cos 72^{\circ}+~E_{B}\cos 72^{\circ}~\)

\( E_{y~net}~=~E~\cos 36^{\circ}~-~E~\cos 36^{\circ}~-~E~\cos 72^{\circ}+~E~\cos 72^{\circ}~\)

\( E_{y~net}~=~0 \)

Therefore, the total electric field at point O is given by,
\(E_{T} = E_{x~net}+ E_{y~net}\)
\(E_{T} = 0 \)


Thus, as shown 5 vectors of equal magnitude at an angle of \(72^{\circ}\) with each other will give a null vector as a resultant at the centre.

Therefore, the point O is equidistant from all the charges at the vertices of pentagon. Thus, due to symmetry, the forces due to all the charges are cancelled out. As a result, electric field at O is zero.

Final Answer: \(E_{net}\) at the centre O is zero.

(ii) As we know that the vector sum of electric field due to charge at A and electric field due to other four charges at the center of pentagon should be zero,

\( \vec{E}_{A}+ \vec{E}_{fourcharges}~=~0 \)

Therefore,

\( \vec{E}_{fourcharges} = -\vec{E}_{A}\)

Hence,

\( \left | \vec{E}_{fourcharges} \right | = \left | \vec{E}_{A} \right |\)

When charge \(q\) is removed from A, net electric field at the centre due to remaining charges will be

\( \left | \vec{E}_{fourcharges} \right | = \left | \vec{E}_{A} \right | = \dfrac{1}{4\pi \epsilon _{0}
} \dfrac{q}{r^{2}} \) along
OA.

Final Answer: \( \left | \vec{E}_{A} \right | = \dfrac{1}{4\pi \epsilon _{0}} \dfrac{q}{r^{2}} \) along OA.

(iii) As per the problem, if charge \(q\) at A is replaced by \(-q\), then electric field due to this negative charge is given by,

\(\vec{E}_{-q} = \dfrac{1}{4\pi \epsilon _{0}
} \dfrac{q}{r^{2}} \) along OA.

Hence net electric field at the center
\( \vec{E}_{net} = \vec{E}_{-q}+ \vec{E}_{fourcharges}~\)

\( \vec{E}_{net} = \dfrac{1}{4\pi \epsilon _{0}
} \dfrac{q}{r^{2}}+\dfrac{1}{4\pi \epsilon _{0}
} \dfrac{q}{r^{2}} \)

\( \vec{E}_{net} = \dfrac{1}{4\pi \epsilon _{0}
} \dfrac{2q}{r^{2}} \) along OA.

Final answer: \( \vec{E}_{net} = \dfrac{1}{4\pi \epsilon _{0}} \dfrac{2q}{r^{2}} \) along OA.

(iv)



As we know, from the given diagram the point O is equidistant from all the charges at the end point of pentagon. Thus, due to symmetry, the forces due to all the charges are cancelled out. As a result, electric field at O is zero.

Now, as per the problem, if pentagon is replaced by n-sided regular polygon with charge \(q\) at each of its comers. The charges will be symmetrical about the centre and therefore the net electric field at O is zero.

Final Answer: \(E_{net}\) at the centre will be same i.e. zero.