Define the electric field strength at a point in an electric field. Figure 2 shows a point charge of +46 μC placed 120 mm from a point charge Q. Position P is on the line joining the charges at a distance 66 mm from charge Q. The resultant electric field strength at position P is zero. Calculate the charge Q. Explain, without calculation, whether net work must be done in moving a proton from infinity to position P in Figure 2. A small rubber ball coated with a conducting paint carries a positive charge. The ball is suspended in equilibrium from a vertical wall by an unchanged non-conducting thread of negligible mass. The wall is positively charged and produces a horizontal uniform electric field perpendicular to the wall along the whole of its length. Figure 3 shows that the thread makes an angle of 30° to the wall. The thread breaks. Explain the motion of the ball.

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Define the electric field strength at a point in an electric field - AQA - A-Level Physics - Question 2 - 2018 - Paper 2

Question 2

Define the electric field strength at a point in an electric field. Figure 2 shows a point charge of +46 μC placed 120 mm from a point charge Q. Position P is on ... show full transcript

Worked Solution & Example Answer:Define the electric field strength at a point in an electric field - AQA - A-Level Physics - Question 2 - 2018 - Paper 2

Step 1

Define the electric field strength at a point in an electric field.

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Answer

The electric field strength at a point is defined as the force per unit charge experienced by a small positive test charge placed at that point.

Step 2

Calculate the charge Q.

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Answer

In this scenario, the electric field strengths produced by both charges at point P must be equal and opposite since the resultant electric field strength is zero. The electric field due to a point charge is given by:

$E = \frac{k \cdot |Q|}{r^2}$

Where:

$E$ is the electric field strength
$k$ is Coulomb's constant (approximately $8.99 \times 10^9 \text{ Nm}^2/\text{C}^2$ )
$|Q|$ is the magnitude of the charge that produces the electric field
$r$ is the distance from the charge to the point.

Let $E_Q$ be the electric field due to charge Q and $E_{46}$ be the electric field due to +46 μC. At position P:

$E_Q = E_{46}$

Hence,

$\frac{k \cdot |Q|}{66^2 \times 10^{-6}} = \frac{k \cdot 46 \times 10^{-6}}{120^2 \times 10^{-6}}$

By canceling $k$ and rearranging, we find:

$|Q| = 46 \times \frac{66^2}{120^2} = 46 \times \frac{4356}{14400} \approx 13.93 \, \mu C.$

Step 3

Explain, without calculation, whether net work must be done in moving a proton from infinity to position P in Figure 2.

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Answer

Work must be done on the positive proton because it is at a positive potential when compared to infinity. Since the proton is moving against the repulsive forces from the like charges, work will be required to move it to position P.

Step 4

Explain the motion of the ball.

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Answer

When the thread breaks, the rubber ball will experience an unbalanced force due to the horizontal electric field. Initially, it is in equilibrium, balancing the tensions from the thread. However, after the thread breaks, the only force acting on the ball is the electric force in the horizontal direction, causing it to accelerate horizontally away from the wall.

Define the electric field strength at a point in an electric field - AQA - A-Level Physics - Question 2 - 2018 - Paper 2

Worked Solution & Example Answer:Define the electric field strength at a point in an electric field - AQA - A-Level Physics - Question 2 - 2018 - Paper 2

Define the electric field strength at a point in an electric field.

Calculate the charge Q.

Explain, without calculation, whether net work must be done in moving a proton from infinity to position P in Figure 2.

Explain the motion of the ball.

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