250+ TOP MCQs on Electrostatic Potential due to a System of Charges | Class12 Physics

Physics Question Papers for Schools on “Electrostatic Potential due to a System of Charges”.

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1. Which is the expression for electric potential due to a system of charges from the following?
a) V=(frac {1}{4 pi varepsilon_o} (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))
b) V=1 χ 4πε_o ( (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))
c) V=(frac {1}{8 pi varepsilon_o} (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))
d) V=(frac {1}{4 pi varepsilon_o} (frac {q_1}{r_1} times frac {q_2}{r_2} times frac {q_3}{r_3} times……. frac {q_n}{r_n}))
Answer: a
Clarification: The electric potential at a point due to a system of charges is equal to the algebraic sum of the electric potentials due to individual charges at that point. The expression for electric potential due to a system of charges is given by:
V=(frac {1}{4 pi varepsilon_o} (frac {q_1}{r_1} + frac {q_2}{r_2} + frac {q_3}{r_3} +……. frac {q_n}{r_n}))

2. There are infinite number of charges, each equal to ‘q’, which are placed along the X-axis at points x = 1, x = 4, x = 16, x = 64……..Then, determine the electric potential due to this system of charges at the point x = 0.
a) V=(frac {16q}{3pi varepsilon_o})
b) V=(frac {3q}{4pi varepsilon_o})
c) V=(frac {4q}{8pi varepsilon_o})
d) V=(frac {3q}{16pi varepsilon_o})
Answer: d
Clarification: We know electric potential(V) due to charge ‘q’ is V = (frac {kq}{r}).
Electric potential (V) is a scalar quantity, so, the total potential at x = 0 is the sum of all the individual charges
V=[((frac {kq}{1}) + (frac {kq}{4}) + (frac {kq}{16}) + (frac {kq}{64}) ) + ……..]
V=kq((frac {1}{(1-frac{1}{4})})) → sum of infinite G.P=(frac {a}{(1-r)})
V=((frac {q}{4pi varepsilon_o}) times (frac {3}{4}))
V=(frac {3q}{16pi varepsilon_o})

3. Two equal and opposite charges Q1 = 2 C and Q2 = -2C are placed at a distance of 6m from each other. What is the potential at the midpoint between the two charges?
a) 2 V
b) 0 V
c) 1 V
d) 3 V
Answer: b
Clarification: Electric potential is a scalar quantity. So, the total potential will be the sum of all the individual charges.
V₁ = (frac {kQ}{r} = frac {2k}{3})
V₂ = (frac {kQ}{r} = frac {-2k}{3})
Total potential (V) = V₁ + V₂
V = k [ ((frac {2}{3}) + (frac {-2}{3})) ]
V = 0 V
Therefore, the potential at the midpoint between the two charges is zero volt.

4. Electric field and electric field intensity are related to electric potential.
a) True
b) False
Answer: a
Clarification: Yes, both are related to electric potential. The relationship between electric potential and electric field is a differential → electric field (E) is the gradient of electric potential (V) in the x direction. Thus, as the charge moves in the x direction, the rate of its change in potential is the value of the electric field. Then, electric field can be defined as the negative of the rate of derivative of potential difference.

5. The 3 charges → (frac {q}{2}), -q, (frac {q}{2}) are placed along the x axis at x = 0, x = r, x = 2 respectively. Find the resultant potential at a point A located at a distance y from charge q such that r << y.
a) V=(frac {4qr^2}{4pi varepsilon_o y})
b) V=(frac {qr^3}{4pi varepsilon_o y^3})
c) V=(frac {qr^2}{4pi varepsilon_o y^3})
d) V=(frac {qr^3}{4pi varepsilon_o y^2})
Answer: c
Clarification: Electric potential at point A can be given by:
V=(frac {1}{4pi varepsilon_o}[(frac {(frac {q}{2})}{y+r}) – frac {q}{y} + frac {(frac {q}{2})}{(y-r)}])
V=(frac {(qr^2)}{4pi varepsilon_o y(y^2 – r^2)})
V=(frac {qr^2}{4pi varepsilon_o y^3}) (Since y >> r)

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