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Chapter 2: Problem 14

Two tiny spheres carrying charges \(1.5 \mu \mathrm{C}\) and \(2.5 \mu \mathrm{C}\) are located \(30 \mathrm{~cm}\) apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point \(10 \mathrm{~cm}\) from this midpoint in a plane normal to the line and passing through the mid-point.

Short Answer

Expert verified
The electric potential at the mid-point (a) will be obtained using the step 2 formula, which will be non-zero. The electric field will be zero, as fields due to both charges cancel each other out. For the point 10 cm away (b), both the electric potential and the electric field will be non-zero, calculated using step 3 formula and considering the directions.

Step by step solution

01

Identifying Variables

Two charges, \(q_1=1.5 \muC\) and \(q_2=2.5 \muC\), are given. Assume the distance between them to be \(d=30cm\). We need to find the electric potential and field at (a) the mid-point and (b) at a point 10 cm away from the mid-point.
02

Calculate Mid-point Potential and Electric Field

For case (a), the mid-point of the two charges has equal distances from both charges. The electric potential \(V\) at a point due to a point charge is given by \(V = \frac{kq}{r}\), where \(k\) is Coulomb's constant and \(r\) is distance from the charge. Both charges will contribute to the electric potential with the same magnitude, since the distances are equal, but the directions will be opposite. Fields due to both charges will cancel each other and total electric field at the mid-point will be zero. Hence, we can calculate potential \(V=2*\frac{k*q}{d/2}\), since the potential is scalar.
03

Calculate Potential and Electric Field at a Point 10 cm Away

For case (b), the point is 10 cm away from the mid-point in a normal plane. If we take positive direction along the line from smaller to larger charge, electric fields from charges will add up, instead of cancelling out like step 2. We can calculate these using \(E = \frac{k*q}{r^2}\), and adding up the electric fields obtained. For potential, use \(V = \frac{k*q}{r}\) again, for each charge with corresponding distance.

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Key Concepts

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

Electric Potential
Electric potential is an important concept to understand when dealing with electric fields and charges. It represents the work done in bringing a unit charge from infinity to a particular point in the field without any acceleration. This work is measured in volts (V). Electric potential is a scalar quantity, meaning it only has magnitude and no direction.
In the case of point charges, the potential at a distance, due to a single point charge, is calculated using the formula: \[ V = \frac{kq}{r} \] where:
  • \(V\) is the electric potential,
  • \(k\) is Coulomb's constant \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\),
  • \(q\) is the point charge, and
  • \(r\) is the distance from the charge.
In the given exercise, at the midpoint between two charges, both will contribute equally to the electric potential.
Because potential is a scalar, their individual potentials simply add up to give the total potential at a point.
Coulomb's Law
Coulomb's law describes the electrostatic interaction between electrically charged particles. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is given by: \[ F = \frac{k |q_1 q_2|}{r^2} \] where:
  • \(F\) is the magnitude of the force between the charges,
  • \(q_1\) and \(q_2\) are the quantities of the charges,
  • \(r\) is the distance separating the charges, and
  • \(k\) is Coulomb's constant.
In simpler words, if the distance between charges increases, the force decreases, showing the inverse square nature of this relationship. This fundamental law helps to calculate both the electric force and the resulting fields when multiple charges are involved.
When we use Coulomb's law to solve electric field problems like in the exercise, it is important to remember that the electric field's direction will depend on the sign of the charges.
Point Charge
A point charge is an idealized model representing a charged object. The charge is concentrated at a single point in space, allowing us to simplify many calculations in electrostatics.
This model is useful because it assumes that all the charge's effects can be concentrated at one point. As a result, the complexities of the physical size and shape of an actual charged body do not complicate our calculations.
In practical applications:
  • Point charges are often used to simplify the study of forces and fields in physics.
  • They allow us to use straightforward formulas like those for electric potential and field.
In the exercise, treating the two small charged spheres as point charges simplifies finding electric potential and fields at specified points. Due to their minimized size in comparison to their separation distance, it's valid to treat them as point charges, allowing one to use the equations derived for point charges for easier computations.

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Most popular questions from this chapter

A spherical capacitor has an inner sphere of radius \(12 \mathrm{~cm}\) and an outer sphere of radius \(13 \mathrm{~cm} .\) The outer sphere is earthed and the inner sphere is given a charge of \(2.5 \mu \mathrm{C}\). The space between the concentric spheres is filled with a liquid of dielectric constant 32 . (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere? (c) Compare the capacitance of this capacitor with that of an isolated sphere of radius \(12 \mathrm{~cm}\). Explain why the latter is much smaller.

A parallel plate capacitor is to be designed with a voltage rating \(1 \mathrm{kV}\), using a material of dielectric constant 3 and dielectric strength about \(10^{7} \mathrm{Vm}^{-1}\). (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say \(10 \%\) of the dielectric strength. What minimum area of the plates is required to have a capacitance of \(50 \mathrm{pF}\) ?

A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show that the capacitance of a spherical capacitor is given by \(C=\frac{4 \pi \varepsilon_{0} r_{1} r_{2}}{r_{1}-r_{2}}\) where \(r_{1}\) and \(r_{2}\) are the radii of outer and inner spheres, respectively.

In a parallel plate capacitor with air between the plates, each plate has an area of \(6 \times 10^{-3} \mathrm{~m}^{2}\) and the distance between the plates is \(3 \mathrm{~mm}\). Calculate the capacitance of the capacitor. If this capacitor is connected to a \(100 \mathrm{~V}\) supply, what is the charge on each plate of the capacitor?

Two charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and \((0,0, a)\). respectively. (a) What is the electrostatic potential at the points \((0,0, z)\) and \((x, y, 0) ?\) (b) Obtain the dependence of potential on the distance \(r\) of a point from the origin when \(r / a \gg 1\). (c) How much work is done in moving a small test charge from the point \((5,0,0)\) to \((-7,0,0)\) along the \(x\) -axis? Does the answer change if the path of the test charge between the same points is not along the \(x\) -axis?
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