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Define electric field strength and give its unit of measurement - Leaving Cert Physics - Question 8 - 2007

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Define electric field strength and give its unit of measurement. Describe how an electric field pattern may be demonstrated in the laboratory. The dome of a Van de... show full transcript

Worked Solution & Example Answer:Define electric field strength and give its unit of measurement - Leaving Cert Physics - Question 8 - 2007

Step 1

Define electric field strength and give its unit of measurement.

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Answer

Electric field strength, denoted by E, is defined as the force F exerted per unit charge q placed in the field. Mathematically, this can be expressed as:

E=FqE = \frac{F}{q}

The unit of measurement for electric field strength is Newtons per Coulomb (N/C) or Volts per meter (V/m).

Step 2

Describe how an electric field pattern may be demonstrated in the laboratory.

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Answer

To demonstrate an electric field pattern, the following apparatus can be used:

  • Apparatus: Oil, metal plates, conductive spherometer, H.T. generator.
  • Procedure: Switch on the power, ensuring that the metal plates are aligned appropriately. The movement of small particles (such as soap bubbles) indicates the electric field pattern, helping to visualize the shape and direction of the electric field.

Step 3

Calculate: (i) the electric field strength at a point 7 cm from the dome.

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Answer

To calculate the electric field strength at a distance of 7 cm from the dome, we can use the formula:

E=14πϵ0Qr2E = \frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}

where:

  • Q=4CQ = 4 C (charge of the dome)
  • r=0.07mr = 0.07 m (distance from the dome surface)
  • ϵ0=8.9×1012 F m1\epsilon_0 = 8.9 \times 10^{-12} \text{ F m}^{-1} (permittivity of free space)

Substituting the values, we calculate:

E=14π(8.9×1012)4(0.07)2E = \frac{1}{4\pi (8.9 \times 10^{-12})} \frac{4}{(0.07)^2}

This results in:

E7.39×1011 N/CE \approx 7.39 \times 10^{11} \text{ N/C}

Step 4

Calculate: (ii) the electrostatic force exerted on the 5 μC point charge.

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Answer

The electrostatic force can be computed using Coulomb's Law:

F=EimesqF = E imes q

where:

  • E7.39×1011 N/CE \approx 7.39 \times 10^{11} \text{ N/C} (from part i)
  • q=5μC=5×106Cq = 5 \mu C = 5 \times 10^{-6} C

Therefore, substituting the values we get:

F=(7.39×1011)×(5×106)3.69×106 NF = (7.39 \times 10^{11}) \times (5 \times 10^{-6}) \approx 3.69 \times 10^{6} \text{ N}

Step 5

Explain why all the charge resides on the surface of a Van de Graff generator's dome.

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Answer

The charge resides on the surface of the dome because like charges repel each other and tend to move as far away from each other as possible. This results in a distribution of charge that is maximally spaced apart, thus accumulating on the outer surface of the dome.

Step 6

Describe an experiment to demonstrate that total charge resides on the outside of a conductor.

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Answer

An experiment can be conducted using a metal can, gold leaf electroscope, and a food plane. The following steps can demonstrate this effect:

  1. Charge the metal can and observe the behavior of the gold leaf electroscope.
  2. Use an insulated wire to connect the can and the electroscope. After charging, note that the leaves diverge, indicating the presence of charge.
  3. Now, touch the inside of the can with your finger briefly. As the charge will redistribute, it can be observed that the leaves of the electroscope return to their original position when disconnected, confirming charge resides on the outside.

Step 7

Give an application of this effect.

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Answer

One application of this effect is in the design of coaxial cables, commonly used for TV signal transmission. These cables utilize the property that the electric field exists only outside the conductors, preventing interference and allowing for efficient signal transmission.

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