Two particles with charges Q1 = +q and Q2 = -3q are fixed in place with a separation 2d. The coordinate of point A is (d, d) and point B is the midpoint between the two charges. Ex... Two particles with charges Q1 = +q and Q2 = -3q are fixed in place with a separation 2d. The coordinate of point A is (d, d) and point B is the midpoint between the two charges. Express your answers in terms of Coulomb constant k, q, and d. (a) Calculate: (i) the electric potential due to the charges at point A, (ii) the electric potential due to the charges at point B, (iii) the change in electric potential energy when a charged particle of charge +2q is moved from point A to point B. (b) Find the x-coordinate between the two charges Q1 and Q2 such that the electric potential due to the two charges is zero.
Understand the Problem
The question involves calculating electric potential and potential energy due to two point charges. Part (a) requires calculating the electric potential at two specific points (A and B) and the change in electric potential energy when a charge is moved between them. Part (b) asks for the x-coordinate where the electric potential due to the two charges is zero. All answers should be expressed in terms of Coulomb constant k, charge q, and distance d.
Answer
(a) (i) $V_A = -\sqrt{2}k\frac{q}{d}$ (a) (ii) $V_B = -2k\frac{q}{d}$ (a) (iii) $\Delta U = (-4 + 2\sqrt{2})k\frac{q^2}{d}$ (b) $x = \frac{d}{2}$
Answer for screen readers
(a) (i) $V_A = -\sqrt{2}k\frac{q}{d}$ (a) (ii) $V_B = -2k\frac{q}{d}$ (a) (iii) $\Delta U = (-4 + 2\sqrt{2})k\frac{q^2}{d}$ (b) $x = \frac{d}{2}$
Steps to Solve
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Calculate the electric potential at point A
- Find the distance from $Q_1$ to point A: $r_1 = \sqrt{(d-0)^2 + (d-0)^2} = \sqrt{2d^2} = d\sqrt{2}$
- Find the distance from $Q_2$ to point A: $r_2 = \sqrt{(d-2d)^2 + (d-0)^2} = \sqrt{(-d)^2 + d^2} = \sqrt{2d^2} = d\sqrt{2}$
- Calculate the electric potential $V_A$ at point A due to both charges: $V_A = k\frac{Q_1}{r_1} + k\frac{Q_2}{r_2} = k\frac{q}{d\sqrt{2}} + k\frac{-3q}{d\sqrt{2}} = k\frac{-2q}{d\sqrt{2}} = -\sqrt{2}k\frac{q}{d}$
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Calculate the electric potential at point B
- Point B is the midpoint between the two charges, so its coordinate is $(d, 0)$.
- Find the distance from $Q_1$ to point B: $r_1 = d$
- Find the distance from $Q_2$ to point B: $r_2 = d$
- Calculate the electric potential $V_B$ at point B due to both charges: $V_B = k\frac{Q_1}{r_1} + k\frac{Q_2}{r_2} = k\frac{q}{d} + k\frac{-3q}{d} = k\frac{-2q}{d} = -2k\frac{q}{d}$
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Calculate the change in electric potential energy
- The change in electric potential energy $\Delta U$ when a charge $+2q$ is moved from point A to point B is given by: $\Delta U = (2q)(V_B - V_A) = 2q(-2k\frac{q}{d} - (-\sqrt{2}k\frac{q}{d})) = 2q(-2k\frac{q}{d} + \sqrt{2}k\frac{q}{d}) = (-4 + 2\sqrt{2})k\frac{q^2}{d}$
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Find the x-coordinate where the electric potential is zero
- Let the x-coordinate be $x$. The distance from $Q_1$ to this point is $x$, and the distance from $Q_2$ to this point is $2d - x$.
- Set the electric potential equal to zero: $V = k\frac{Q_1}{x} + k\frac{Q_2}{2d - x} = k\frac{q}{x} + k\frac{-3q}{2d - x} = 0$
- Solve for $x$: $\frac{q}{x} - \frac{3q}{2d - x} = 0$ $\frac{q}{x} = \frac{3q}{2d - x}$ $2d - x = 3x$ $2d = 4x$ $x = \frac{d}{2}$
(a) (i) $V_A = -\sqrt{2}k\frac{q}{d}$ (a) (ii) $V_B = -2k\frac{q}{d}$ (a) (iii) $\Delta U = (-4 + 2\sqrt{2})k\frac{q^2}{d}$ (b) $x = \frac{d}{2}$
More Information
The electric potential is a scalar quantity, meaning it has magnitude but not direction. The electric potential energy is the energy a charged particle has due to its position in an electric field.
Tips
- Forgetting to consider the sign of the charges when calculating the electric potential.
- Incorrectly calculating the distances between the charges and the points where the potential is being calculated.
- Making algebraic errors when solving for the x-coordinate where the electric potential is zero.
- Confusing electric potential with electric potential energy.
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