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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 when moving a charge from one point to another in an electric field. Its unit is the volt (V), where 1 volt = 1 joule per coulomb (J/C).

Step 2

Define electric field strength.

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Answer

Electric field strength (E) is defined as the force (F) experienced by a unit positive charge (Q) placed in the field. Mathematically, it is expressed as:

E=FQE = \frac{F}{Q}

Its unit is newtons per coulomb (N/C).

Step 3

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

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To charge an insulated spherical conductor positively by induction:

  1. Bring a negatively charged rod close to the conductor without touching it.
  2. The presence of the negative charge induces a positive charge on the near side of the conductor, while the far side becomes negatively charged.
  3. Ground the far side of the conductor to allow the negative charges to flow away.
  4. Remove the ground connection and then the negatively charged rod. The conductor now retains a net positive charge.

Step 4

Draw the electric field around the charged conductor.

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The electric field lines around a positively charged spherical conductor radiate outward, away from the surface of the sphere. The lines are denser near the surface, indicating a stronger field, and they taper off as they move further away.

Step 5

Calculate the charge on the conductor.

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Answer

To calculate the charge on the conductor, we use the formula for electric field strength:

E=Q4 π ϵ0⋅r2E = \frac{Q}{4 \, \pi \, \epsilon_0 \cdot r^2}

Given:

  • Electric field strength, E=2.3 N C−1E = 2.3 \text{ N C}^{-1}
  • Distance from the surface, r=0.05 mr = 0.05 \text{ m}:

Rearranging gives:

Q=E⋅4πϵ0⋅r2Q = E \cdot 4 \pi \epsilon_0 \cdot r^2 Substituting  ϵ0=8.85×10−12 C2/N m2\,\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2,

Q=2.3⋅4π(8.85×10−12)(0.052)Q = 2.3 \cdot 4 \pi (8.85 \times 10^{-12})(0.05^2)

Calculating this gives:

Q=3.1×10−12 CQ = 3.1 \times 10^{-12} \text{ C}

Step 6

Explain how point discharge occurs.

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Answer

Point discharge occurs when charge accumulates at a pointed electrode. The electric field around the point becomes extremely strong, ionizing the air at that point. The ionized air allows charges to escape into the atmosphere, creating a stream of ions. Consequently, point discharge creates a visible spark or corona effect due to the rapid movement of ions.

Step 7

Describe how point discharge can be demonstrated in the laboratory.

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Answer

Point discharge can be demonstrated in the laboratory using:

  1. Charged Points: A pointed conductor charged positively can be used.
  2. Candle at Point: Placing a candle near the point while it is charged will show the light flickering due to air ionization.
  3. Flame Blown Away from the Point: The flame can be blown away from the point, indicating the discharge occurring away from the point.

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