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An object with charge \(q = -6.00 \times 10^{-9}\) C is placed in a region of uniform electric field and is released from rest at point \(A\). After the charge has moved to point \(B\), 0.500 m to the right, it has kinetic energy \(3.00 \times 10^{-7}\) J. (a) If the electric potential at point \(A\) is \(+\)30.0 V, what is the electric potential at point \(B\)? (b) What are the magnitude and direction of the electric field?

Short Answer

Expert verified
(a) The electric potential at point B is 80.0 V. (b) The electric field magnitude is 100 V/m, directed left.

Step by step solution

01

Understanding the Relationship Between Kinetic Energy and Electric Potential

The change in kinetic energy of the charge as it moves from point A to B is equal to the work done by the electric field on the charge. The work done by the electric field is also equal to the change in electric potential energy. Hence, we use the relation: \[ \Delta U = -\Delta K = q( V_B - V_A) \] where \( \Delta K \) is the change in kinetic energy, \( q \) is the charge, and \( V_A \) and \( V_B \) are the electric potentials at points A and B, respectively. Given \( \Delta K = 3.00 \times 10^{-7} \) J and \( q = -6.00 \times 10^{-9} \) C, we need to find \( V_B \).
02

Calculating the Electric Potential at Point B

We first rearrange the equation:\[ q (V_B - V_A) = - \Delta K \]Substitute the respective values, with \( V_A = 30.0 \) V:\[ -6.00 \times 10^{-9} (V_B - 30.0) = -3.00 \times 10^{-7} \]Solving for \( V_B \), we get:\[ V_B - 30.0 = \frac{3.00 \times 10^{-7}}{6.00 \times 10^{-9}} \]\[ V_B - 30.0 = 50 \]\[ V_B = 80.0 \] V.
03

Understanding Electric Field and Potential Difference

The electric field \( E \) is related to the potential difference \( \Delta V \) between two points separated by a distance \( d \) by the formula:\[ E = \frac{\Delta V}{d} \]where \( \Delta V = V_B - V_A \) and \( d = 0.500 \) m, the distance between points A and B. We know \( V_B = 80.0 \) V and \( V_A = 30.0 \) V.
04

Calculating the Magnitude of the Electric Field

Substituting the values we get:\[ E = \frac{80.0 - 30.0}{0.500} \]\[ E = \frac{50.0}{0.500} \]\[ E = 100 \] V/m.
05

Determining the Direction of the Electric Field

The object has a negative charge and is moving to the right from point A to B under the influence of the electric field. A negative charge moves in the direction opposite to that of the electric field. Therefore, the electric field points to the left.

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Key Concepts

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

Work Done by Electric Field
When a charged particle moves in an electric field, work is done on it by the field. This work done by the electric field results in a change in the particle's kinetic energy. In simple terms, when an object with an electric charge moves from one point to another within an electric field, the force exerted by this field can speed up or slow down the object. This interaction changes the object's kinetic energy.

The relationship is expressed in the formula: - \[ W = - riangle K = q( V_B - V_A) \] - where:
  • \( W \) is the work done by the electric field
  • \( riangle K \) is the change in kinetic energy
  • \( q \) is the charge of the object
  • \( V_B \) and \( V_A \) are the electric potentials at points B and A
When an object moves in the direction of an electric field, work is done which is related to the potential difference between the two points.
Electric Field Direction
The direction of the electric field is crucial in understanding how charges move. In this scenario, the charge is negative. Negative charges move opposite to the direction of the electric field. So, if a negative charge moves from point A to point B, it suggests that the electric field direction is opposite to the charge's movement.

Therefore, for our example where the charge moves to the right, the electric field must be pointing to the left. This is because:
  • Electric fields point away from positive charges and toward negative charges.
  • A negative charge moves against the direction of the electric field, so if it moves right, the field is left.
Understanding this concept helps in predicting the behavior of charged particles in electric fields.
Change in Kinetic Energy
The change in kinetic energy of a charge is key to understanding the effects of an electric field. Initially, the charge starts from rest, meaning its initial kinetic energy is zero. When the charge moves through the electric field to another point, its kinetic energy changes due to the work done by the field.

For a better understanding, consider the formula: \[ riangle K = K_B - K_A \] where:
  • \( K_A \) is the initial kinetic energy (which is zero since it starts from rest)
  • \( K_B \) is the final kinetic energy (given as \( 3.00 \times 10^{-7} \) J)
The change indicates how much the charge has accelerated due to the electric field's work.
Electric Potential Difference
Electric potential difference, also known as voltage, is the difference in electric potential energy between two points in a field, per unit charge. It tells us how much work is needed to move a charge from one point to another.

Given the potential at one point \( V_A = 30.0 \) V, we need to calculate the potential at another point \( V_B \). Utilizing the work-energy principle, we use: \[ q(V_B - V_A) = - riangle K \] With the charge \( q = -6.00 \times 10^{-9} \) C and the kinetic energy change \( \triangle K = 3.00 \times 10^{-7} \) J, we solved for \( V_B \) as 80.0 V.

This potential difference affects how charges move within the field and is vital for understanding how fields do work on charges.

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Most popular questions from this chapter

A metal sphere with radius \(r_a\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(r_b\) . There is charge \(+\)q on the inner sphere and charge \(-\)q on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(r < r_a ;\) (ii) \(r_a < r < r_b ;\) (iii) \(r > r_b .\) (\(Hint\): The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$V_{ab} = \frac{q} {4\pi\epsilon_0} ( \frac{1} {r_a} - \frac{1} {r_b} )$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude $$E(r) = \frac {V_{ab}} {(1/r_a - 1/r_b)}\frac {1} {r_2}$$ (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r > r_b\) . (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q\). Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 3.75 kV? (b) What is the potential of the sphere 's surface relative to infinity?

A uniformly charged, thin ring has radius 15.0 cm and total charge \(+\)24.0 nC. An electron is placed on the ring's axis a distance 30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.

Charge \(Q =\) 5.00 mC is distributed uniformly over the volume of an insulating sphere that has radius \(R =\) 12.0 cm. A small sphere with charge \(q = +\)3.00 \(\mu\)C and mass 6.00 \(\times 10^{-5}\) kg is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 cm of the surface of the large sphere?

A thin spherical shell with radius \(R_1 =\) 3.00 cm is concentric with a larger thin spherical shell with radius \(R_2 =\) 5.00 cm. Both shells are made of insulating material. The smaller shell has charge \(q_1 = +\)6.00 nC distributed uniformly over its surface, and the larger shell has charge \(q_2 = -\)9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) \(r =\) 0; (ii) \(r =\) 4.00 cm; (iii) \(r =\) 6.00 cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

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