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In a thunderstorm different parts of a cloud become positively and negatively charged - Leaving Cert Physics - Question 7 - 2019

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In a thunderstorm different parts of a cloud become positively and negatively charged. There is a large electric field and a large potential difference between diffe... show full transcript

Worked Solution & Example Answer:In a thunderstorm different parts of a cloud become positively and negatively charged - Leaving Cert Physics - Question 7 - 2019

Step 1

What is meant by potential difference? State its unit.

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Answer

Potential difference refers to the work done per unit charge in moving a positive test charge between two points in an electric field. Its unit is the volt (V), where 1 volt is equivalent to 1 joule per coulomb (1 V = 1 J C⁻¹).

Step 2

Define electric field strength.

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Answer

Electric field strength is defined as the force experienced by a unit positive charge placed in an electric field. It is given mathematically by the formula: E=FQE = \frac{F}{Q} where E is the electric field strength, F is the force, and Q is the charge. Its unit is also volts per meter (V/m).

Step 3

Describe how an insulated spherical conductor can be charged positively by induction.

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Answer

  1. Negatively charged rod close to: Bring a negatively charged rod close to the insulated spherical conductor without touching it.
  2. Earth sphere: This will cause electrons in the conductor to be repelled away from the rod, creating a positive charge on the side of the sphere nearest to the rod.
  3. Remove earth: If the conductor is then grounded (earthed), electrons will flow away from the conductor into the ground, leaving the conductor positively charged when the grounding connection is removed.

Step 4

Draw the electric field around the charged conductor.

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Answer

The electric field lines around a positively charged spherical conductor radiate outward from the surface and are radially symmetric. The field lines demonstrate that the strength of the electric field decreases as one moves further away from the charge.

Step 5

Calculate the charge on the conductor.

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Answer

Using the electric field intensity formula: E=Q4πε0r2E = \frac{Q}{4\pi\varepsilon_0 r^2} Where:

  • E = 2.3 N C⁻¹,
  • r = 5 cm = 0.05 m,
  • \varepsilon_0 = 8.85 \times 10^{-12} C^2/(N \cdot m^2).

Rearranging gives: Q=E4πε0r2Q = E \cdot 4\pi\varepsilon_0 r^2
Q=2.3×(4π×8.85×1012×(0.05)2)=3.1×1012CQ = 2.3 \times (4 \pi \times 8.85 \times 10^{-12} \times (0.05)^2) = 3.1 \times 10^{-12} C.

Step 6

Explain how point discharge occurs.

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Answer

Point discharge occurs when the electric field strength at a sharp point becomes large enough to ionize the surrounding air. The process involves:

  1. Charge accumulates at a point: The sharp point collects charge, increasing the field strength.
  2. Air around point is ionised: When the field strength exceeds the dielectric strength of air, it ionizes the air, allowing current to flow.
  3. Opposite charges attracted / like charges repelled: This ionization process allows surrounding air molecules to be attracted or repelled, leading to visible discharges such as sparks.

Step 7

Describe how point discharge can be demonstrated in the laboratory.

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Answer

  1. Charged point: Set up a charged pointed conductor (e.g., a pointed rod connected to a high-voltage source).
  2. Arcing can occur: Bring the conductor close to a grounded plate, and observe small sparks as the air around the point ionizes.
  3. Flame is blown away from the point: Alternatively, bring a lightly dampened piece of cloth near the point; the discharge can cause it to ignite, demonstrating point discharge.

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