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Question: An infinite number of electric charges each equal to \(2\,nano\, coulombs \) in magnitude are placed...

An infinite number of electric charges each equal to 2nanocoulombs2\,nano\, coulombs in magnitude are placed along x-axis at x=1cmx = 1\,cm, x=3cmx = 3\,cm, x=9cmx = 9\,cm, x=27cmx = 27\,cm …. And so on. In this setup if the consecutive charges have opposite sign, then the electric potential at x=0x = 0 is
A. 1250V1250\,V
B. 1350V1350\,V
C. 2700V2700\,V
D. 2500V2500\,V

Explanation

Solution

In this question, we need to find the electric potential at x=0x = 0. We will use the formula of the electric potential at any point around a point charge QQ. And evaluate the rr, then we will apply the values and evaluate to determine the required solution.

Complete step by step answer:
The electric potential energy is the amount of work needed to move a unit of electric charge from a reference point to a specific point in an electric field without producing acceleration.Now, we know that the formula of the electric potential at any point around a point charge QQ is given by,
V=k[Qr]V = k\left[ {\dfrac{Q}{r}} \right]
Where VV is the electric potential energy,QQ is a point charge, rr is the distance between any point around the charge to the point charge and kk is Coulomb constant; k=9.0×109Nk = 9.0 \times {10^9}\,N.
V=kQ(1r1+1r2+1r3+1r4...)V= kQ\left( {\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}} + \dfrac{1}{{{r_3}}} + \dfrac{1}{{{r_4}}}...} \right)
V=kQ(1+13+19+127...)\Rightarrow V = kQ\left( {1 + \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}}...} \right)
V=kQ(1+13+132+133...)\Rightarrow V = kQ\left( {1 + \dfrac{1}{3} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{3^3}}}...} \right)
V=kQ(1113)\Rightarrow V = kQ\left( {\dfrac{1}{{1 - \dfrac{1}{3}}}} \right)
V=kQ×32\Rightarrow V = kQ \times \dfrac{3}{2}
Therefore at x=0x = 0, the electric potential is,
V1=9×109×2×109×32{V_1} = 9 \times {10^9} \times 2 \times {10^{ - 9}} \times \dfrac{3}{2}
V1=18×32\Rightarrow{V_1} = 18 \times \dfrac{3}{2}
V1=2700V\Rightarrow {V_1} = 2700\,V
It is also given that the consecutive charges have opposite signs.
Therefore, the electric potential =V12 = \dfrac{{{V_1}}}{2}
By substituting V1=2700V{V_1} = 2700\,V,
27002\Rightarrow\dfrac{{2700}}{2}
1350V\Rightarrow 1350\,V

Hence, option B is the correct answer.

Note: In this question, it is important to note here that, we may think V1=2700V{V_1} = 2700\,V is the required solution and stop upto it but in the question it is also given the consecutive charges have opposite signs so we need divide by 22 in order to get the required electric potential. So, read the given carefully while solving these types of questions.