smc-4283-30-Forces on charges

PHYSICS, M4 2023 VCE 1

Some physics students are conducting an experiment investigating both electrostatic and gravitational forces. They suspend two equally charged balls, each of mass 4.0 g, from light, non-conducting strings suspended from a low ceiling.

The charged balls repel each other with the strings at an angle of 60°, as shown in Figure 1.

There are three forces acting on each ball:

- a tension force, \(T\)
- a gravitational force, \(F_{g}\)
- an electrostatic force, \(F_{E}\).

On Figure 1, using the labels \(T, F_{g}\) and \(F_{E}\), draw each of the three forces acting on each of the charged balls. (3 marks)

--- 0 WORK AREA LINES (style=lined) ---
Show that the tension force, \(T\), in each string is \(4.5 \times 10^{-2} \text{ N}\). Use \(g=9.8 \text{ N kg}^{-1}\).
Show your working. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---
Calculate the magnitude of the electrostatic force, \(F_{ E }\). Show your working. (2 marks)

--- 6 WORK AREA LINES (style=lined) ---

Show Answers Only

b. System is in an equilibrium state:

→ The sum of the forces must add to zero as seen in the triangle below.

→ \(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\)

\(\sin\,60^{\circ}\)	\(=\dfrac{3.92 \times 10^{-2}}{T}\)
\(T\)	\(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}\)
	\(=4.5 \times 10^{-2}\ \text{N}\)

c. Using the same triangle as part (b):

\(\tan\,60^{\circ}\)	\(=\dfrac{3.92 \times 10^{-2}}{F_E}\)
\(F_E\)	\(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}\)
	\(=2.3 \times 10^{-2}\ \text{N}\)

Show Worked Solution

b. System is in an equilibrium state:

→ The sum of the forces must add to zero as seen in the triangle below.

→ \(F_g = 9.8 \times 4 \times 10^{-3} = 3.92 \times 10^{-2}\ \text{N}\)

\(\sin\,60^{\circ}\)	\(=\dfrac{3.92 \times 10^{-2}}{T}\)
\(T\)	\(=\dfrac{3.92 \times 10^{-2}}{\sin\,60^{\circ}}\)
	\(=4.5 \times 10^{-2}\ \text{N}\)

c. Using the same triangle as part (b):

\(\tan\,60^{\circ}\)	\(=\dfrac{3.92 \times 10^{-2}}{F_E}\)
\(F_E\)	\(=\dfrac{3.92 \times 10^{-2}}{\tan\,60^{\circ}}\)
	\(=2.3 \times 10^{-2}\ \text{N}\)

♦ Mean mark 41%.

PHYSICS, M4 2023 VCE 2 MC

The diagram below shows two charges, \(Q_1\) and \(Q_2\), separated by a distance, \(d\).

There is a force, \(F\), acting between the two charges.

Which one of the following is closest to the magnitude of the force acting between the two charges if both \(d\) and the charge on \(Q_1\) are halved?

\(\dfrac{F}{4}\)
\(F\)
\(2 F\)
\(4 F\)

Show Answers Only

\(C\)

Show Worked Solution

\(\text{Initially,}\ \ F=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{Q_1Q_2}{d^2}\)

\(\text{After}\ \ Q_1\ \ \Rightarrow \ \ \dfrac{Q_1}{2}, \ \text{and}\ \ d\ \ \Rightarrow \ \ \dfrac{d}{2}:\)

\(F_{\text{new}}\)	\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{\frac{Q_1}{2}Q_2}{(\frac{d}{2})^2}\)
	\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{\frac{Q_1}{2}Q_2}{\frac{d^2}{4}}\)
	\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{4Q_1Q_2}{2d^2}\)
	\(=\dfrac{1}{4 \pi \varepsilon_0} \times \dfrac{2Q_1Q_2}{d^2}\)
	\(=2F\)

\(\Rightarrow C\)

♦ Mean mark 55%.

PHYSICS, M4 2019 VCE 3*

Three charges \((- \text{Q} ,+2 \text{Q} ,-2 \text{Q})\) are placed at the vertices of an isosceles triangle, as shown below.

Draw a force vector diagram to show the direction of the net force on the charge \(- \text{Q}\) ? (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

Show Worked Solution

→ \(- \text{Q}\) will be attracted to \(+2 \text{Q}\) charge and repelled from the \(-2 \text{Q}\) charge.

→ This results in a force vector diagram as seen below.

PHYSICS, M4 2021 VCE 5b

Figure 5 shows a stationary electron \(\left( \text{e} ^{-}\right)\) in a uniform magnetic field between two parallel plates. The plates are separated by a distance of 6.0 × 10\(^{-3}\) m, and they are connected to a 200 V power supply and a switch. Initially, the plates are uncharged. Assume that gravitational effects on the electron are negligible.

The switch is now closed.

Determine the magnitude and the direction of any electric force now acting on the electron. Show your working. (3 marks)

Show Answers Only

\(F=5.34 \times 10^{-15}\ \text{N towards the bottom plate.}\)

Show Worked Solution

→ Determine the magnitude of the electric force on the electron:

\(F=qE = \dfrac{qV}{d}=\dfrac{1.602 \times 10^{-19} \times 200}{6.0 \times 10^{-3}}=5.34 \times 10^{-15}\ \text{N}\)

→ The longer line of the battery cell represents the positive terminal, hence the bottom plate will be the positive plate and the electric force on the electron will be towards the bottom plate.

♦ Mean mark 53%.

PHYSICS, M4 2015 HSC 24b

A part of a cathode ray oscilloscope was represented on a website as shown.

Electrons leave the cathode and are accelerated towards the anode.

Calculate the force on an electron due to the electric field between the cathode and the anode. (2 marks)

Show Answers Only

\(F=4 \times 10^{-14}\) N

Show Worked Solution

Using \(F=qE\) and \(E=\dfrac{V}{d}\):

\(F\)	\(=\dfrac{qV}{d}\)
	\(=\dfrac{1.602 \times 10^{-19} \times 5000}{0.02}\)
	\(=4 \times 10^{-14}\) N

PHYSICS, M4 2017 HSC 8 MC

An electron is fired in a vacuum towards a screen. With no electric field being applied, the electron hits the screen at \(P\). A uniform electric field is turned on and another electron is fired towards the screen from the same location, at the same velocity, striking the screen at point \(Q\).

With the electric field still turned on, a proton is fired towards the screen from the same starting point as the electrons and with the same velocity.

At what point does the proton strike the screen?

\(A\)
\(B\)
\(C\)
\(D\)

Show Answers Only

\(C\)

Show Worked Solution

→ As the proton has an equal but opposite charge to the electron it will experience a force in the opposite direction. Hence it will hit the screen to the right of \(P\).

→ Due to the proton having a greater mass, it will be deflected by a smaller amount and so will hit the screen at \(C\).

\(\Rightarrow C\)

♦♦♦ Mean mark 25%.