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Two point charges, \(+3.40 \mu \mathrm{C}\) and \(-6.10 \mu \mathrm{C}\) , are separated by 1.20 \(\mathrm{m}\) . What is the electric potential midway between them?

Short Answer

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Step by step solution

01

Understand the Concept of Electric Potential

Electric potential at a point due to a point charge is given by the formula \( V = \frac{k \cdot q}{r} \) where \( V \) is the electric potential, \( k \) is the Coulomb's constant \((8.99 \times 10^9 \, \mathrm{Nm^2/C^2})\), \( q \) is the charge, and \( r \) is the distance from the charge to the point of interest. The electric potential is scalar and can be added algebraically.
02

Identify the Contributions

There are two charges: \(+3.40 \mu \mathrm{C}\) and \(-6.10 \mu \mathrm{C}\). Both will contribute to the electric potential at the midpoint. The distance to the midpoint from each charge is half the distance between them, so \( r = \frac{1.20}{2} \mathrm{m} = 0.60 \,\mathrm{m}\).
03

Calculate Potential Due to Each Charge

Using the formula for electric potential, calculate the potential due to each charge at the midpoint:- For the \(+3.40 \mu \mathrm{C}\) charge: \[ V_1 = \frac{8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \cdot 3.40 \times 10^{-6} \,\mathrm{C}}{0.60 \,\mathrm{m}} = 5.0947 \times 10^4 \, \mathrm{V} \]- For the \(-6.10 \mu \mathrm{C}\) charge: \[ V_2 = \frac{8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \cdot (-6.10 \times 10^{-6} \,\mathrm{C})}{0.60 \,\mathrm{m}} = -9.1475 \times 10^4 \, \mathrm{V} \]
04

Sum the Potentials

The total electric potential at the midpoint is the sum of the potentials due to the two charges:\[V = V_1 + V_2 = 5.0947 \times 10^4 \, \mathrm{V} + (-9.1475 \times 10^4 \, \mathrm{V})\]\[V = -4.0528 \times 10^4 \, \mathrm{V}\]
05

Final Result

Thus, the electric potential midway between the two point charges is \(-4.0528 \times 10^4 \, \mathrm{V}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Constant
When it comes to electric potential and electric force, Coulomb's constant is a key player. This constant, denoted by the letter "k," is a proportional factor in the electric force between two point charges. Its value is approximately \( 8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \). This high value reflects the strength of the forces between charged particles.
Coulomb's constant is derived from Coulomb's Law, which quantifies the amount of force between two stationary, electrically charged particles. In the context of electric potential, it is used to calculate the potential energy per unit charge due to another charge.
Using \( k \) in the electric potential formula allows us to find out how much potential energy a test charge would have in the electric field created by another charge. This shows that even a small charge can have a significant potential due to the large multiplying factor of Coulomb's constant.
Point Charge
A point charge is a model of a charged object, used to simplify calculations in electromagnetism and electrostatics. It's an idealized model where the physical size of the charge is so small compared to the distances of interest, that it can be treated as a point.
Unlike distributed charges, which spread over a volume or surface, a point charge is considered to be concentrated at a single location. This makes calculations easier, as the electric field or potential can be calculated as if the entire charge is located at a single point.
In the problem above, the charges given, \(+3.40 \mu \mathrm{C}\) and \(-6.10 \mu \mathrm{C}\), are treated as point charges. This treatment simplifies finding the electric potential at the midpoint, as we only need to consider the straight-line distance between the charges and the point of interest.
Electric Potential Formula
The electric potential formula is essential when calculating the potential at a specific point from a point charge. It is given by \( V = \frac{k \cdot q}{r} \), where \( V \) is the electric potential, \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point of interest.
Electric potential is a scalar quantity. This means it doesn't have direction, unlike electric fields, which are vector quantities. Therefore, the potentials from different charges can simply be added together to find the total potential at a point. This property makes potential calculations more straightforward.
In the exercise, we used this formula to determine the electric potential at the midpoint between the given charges. For each charge, we applied the formula separately and then summed the two results. By doing this, we found the net potential due to the effects of both point charges at the midpoint. Understanding how to use this formula is crucial for problems involving multiple charges and different configurations.

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Most popular questions from this chapter

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.2 \(\mathrm{cm}\) , and the electric field within the capacitor has a magnitude of 2.1 \(\times 10^{6} \mathrm{V} / \mathrm{m}\) . What is the kinetic energy of the electron just as it reaches the positive plate?

The work done by an electric force in moving a charge from point \(A\) to point \(B\) is \(2.70 \times 10^{-3} \mathrm{J}\) . The electric potential difference between the two points is \(V_{A}-V_{B}=50.0 \mathrm{V}\) . What is the charge?

If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude that the electric field can have without breakdown occurring. The dielectric strength of air is \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m},\) and that of neoprene rubber is \(1.2 \times 10^{7} \mathrm{V} / \mathrm{m} .\) A certain air-gap, parallel plate capacitor can store no more than 0.075 \(\mathrm{J}\) of electrical energy before breaking down. How much energy can this capacitor store with- out breaking down after the gap between its plates is filled with neoprene rubber?

The membrane that surrounds a certain type of living cell has a surface area of \(5.0 \times 10^{-9} \mathrm{m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{m} .\) Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to positive ions (charge \(+e ),\) how many such ions are present on the outer surface?

Charges of \(-q\) and \(+2 q\) are fixed in place, with a distance of 2.00 \(\mathrm{m}\) between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance \(L\) from the negative charge, there is at least one spot where the total potential is zero. Find \(L .\)

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