(a) First, we need to find the kinetic energy of the electron after it has been accelerated across the potential difference. The kinetic energy (KE) can be calculated using the formula KE = qV, where q is the charge of the electron and V is the potential difference. The charge of an electron is -1.6 x 10^-19 C and the potential difference is 9 V. So, KE = -1.6 x 10^-19 C * 9 V = -1.44 x 10^-18 J.

Next, we need to find the work done by the electric field to bring the electron to a halt. The work done (W) is equal to the change in kinetic energy, which in this case is the kinetic energy we just calculated. So, W = -1.44 x 10^-18 J.

The work done by the electric field can also be calculated using the formula W = Fd, where F is the force and d is the distance. The force exerted by the electric field on the electron can be calculated using the formula F = Eq, where E is the electric field strength and q is the charge of the electron. So, F = 5.7 N/C * -1.6 x 10^-19 C = -9.12 x 10^-19 N.

Setting the two expressions for the work done equal to each other, we get -1.44 x 10^-18 J = -9.12 x 10^-19 N * d. Solving for d, we find that d = 1.44 x 10^-18 J / 9.12 x 10^-19 N = 1.58 m. So, the electron is able to travel 1.58 m into the region with the electric field before it is brought to a halt.

(b) The electric potential (V) at a point in an electric field can be calculated using the formula V = Ed, where E is the electric field strength and d is the distance from the reference point. In this case, the reference point is where the electron enters the field, so d = 1.58 m. So, V = 5.7 N/C * 1.58 m = 9 V. Therefore, the electric potential at the point the electron stops is 9 V relative to the point where the electron enters the field.

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(a) First, we need to find the kinetic energy of the electron after it has been accelerated across the potential difference. The kinetic energy (KE) can be calculated using the formula KE = qV, where q is the charge of the electron and V is the potential difference. The charge of an electron is -1.6 x 10^-19 C and the potential difference is 9 V. So, KE = -1.6 x 10^-19 C * 9 V = -1.44 x 10^-18 J.

Next, we need to find the work done by the electric field to bring the electron to a halt. The work done (W) is equal to the change in kinetic energy, which in this case is the kinetic energy we just calculated. So, W = -1.44 x 10^-18 J.

The work done by the electric field can also be calculated using the formula W = Fd, where F is the force and d is the distance. The force exerted by the electric field on the electron can be calculated using the formula F = Eq, where E is the electric field strength and q is the charge of the electron. So, F = 5.7 N/C * -1.6 x 10^-19 C = -9.12 x 10^-19 N.

Setting the two expressions for the work done equal to each other, we get -1.44 x 10^-18 J = -9.12 x 10^-19 N * d. Solving for d, we find that d = 1.44 x 10^-18 J / 9.12 x 10^-19 N = 1.58 m. So, the electron is able to travel 1.58 m into the region with the electric field before it is brought to a halt.

(b) The electric potential (V) at a point in an electric field can be calculated using the formula V = Ed, where E is the electric field strength and d is the distance from the reference point. In this case, the reference point is where the electron enters the field, so d = 1.58 m. So, V = 5.7 N/C * 1.58 m = 9 V. Therefore, the electric potential at the point the electron stops is 9 V relative to the point where the electron enters the field.

Knowee AI · Accepted Answer

(a) First, we need to find the kinetic energy of the electron after it has been accelerated across the potential difference. The kinetic energy (KE) can be calculated using the formula KE = qV, where q is the charge of the electron and V is the potential difference. The charge of an electron is -1.6 x 10^-19 C and the potential difference is 9 V. So, KE = -1.6 x 10^-19 C * 9 V = -1.44 x 10^-18 J.

Next, we need to find the work done by the electric field to bring the electron to a halt. The work done (W) is equal to the change in kinetic energy, which in this case is the kinetic energy we just calculated. So, W = -1.44 x 10^-18 J.

The work done by the electric field can also be calculated using the formula W = Fd, where F is the force and d is the distance. The force exerted by the electric field on the electron can be calculated using the formula F = Eq, where E is the electric field strength and q is the charge of the electron. So, F = 5.7 N/C * -1.6 x 10^-19 C = -9.12 x 10^-19 N.

Setting the two expressions for the work done equal to each other, we get -1.44 x 10^-18 J = -9.12 x 10^-19 N * d. Solving for d, we find that d = 1.44 x 10^-18 J / 9.12 x 10^-19 N = 1.58 m. So, the electron is able to travel 1.58 m into the region with the electric field before it is brought to a halt.

(b) The electric potential (V) at a point in an electric field can be calculated using the formula V = Ed, where E is the electric field strength and d is the distance from the reference point. In this case, the reference point is where the electron enters the field, so d = 1.58 m. So, V = 5.7 N/C * 1.58 m = 9 V. Therefore, the electric potential at the point the electron stops is 9 V relative to the point where the electron enters the field.

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