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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 to move a positive charge between two points in an electric field. The unit of potential difference is the volt (V), which is defined as one joule per coulomb (1 V = 1 J/C).

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 the field. It can be expressed mathematically as:

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

Where:

  • EE is the electric field strength (N/C),
  • FF is the force (N), and
  • qq is the charge (C).

Step 3

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

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  1. Place a negatively charged object close to the insulated spherical conductor. This will induce a separation of charge within the conductor.

  2. The negative charges in the conductor will be attracted towards the negatively charged object, leaving a positive charge on the side farthest from it.

  3. Insert an earth connection to the conductor. This allows electrons to flow from the conductor to the ground, thus neutralizing the negative charges.

  4. Remove the earth connection and then the negatively charged object. The spherical conductor will retain a net positive charge.

Step 4

Draw the electric field around the charged conductor.

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Answer

- Radial field lines will radiate outward from the positive charge of the spherical conductor.
- The density of these lines represents the strength of the electric field; closer lines indicate a stronger field. 

Refer to a diagram showing uniformly spaced radial lines extending outwards from the surface of the conductor.

Step 5

Calculate the charge on the conductor.

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Answer

The electric field EE at a distance of r=5extcm=0.05extmr = 5 ext{ cm} = 0.05 ext{ m} from the surface of the spherical conductor is given as 2.3 N/C. We can use the formula:

E=Q4πϵ0r2E = \frac{Q}{4\pi \epsilon_0 r^2}

Rearranging gives:

Q=E4πϵ0r2Q = E \cdot 4\pi \epsilon_0 r^2

Substituting the values:

  • E=2.3extN/CE = 2.3 ext{ N/C}
  • ϵ0=8.85×1012extC2/N m2\epsilon_0 = 8.85 \times 10^{-12} ext{ C}^2/\text{N m}^2
  • r=0.12extmr = 0.12 ext{ m} (diameter 12 cm)

Calculating:

Q=2.3×4π(8.85×1012)(0.05)2=3.1×1012extCQ = 2.3 \times 4\pi (8.85 \times 10^{-12}) (0.05)^2 \\ = 3.1 \times 10^{-12} ext{ C}

Step 6

Explain how point discharge occurs.

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Answer

Point discharge occurs when a charged object, such as a conductor, has a high electric field strength at its point.

  1. Charge accumulates at that point due to its geometry, creating a strong local electric field.

  2. If the air around the point is ionized (meaning enough energy is provided to electrons that they can move freely), this electric field can cause electrons in the air to become free, forming an ionized path.

  3. Once a conductive path is formed, opposite charges can rapidly discharge, neutralizing the accumulated charge.

  4. This can result in phenomena like lightning or corona discharge.

Step 7

Describe how point discharge can be demonstrated in the laboratory.

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Answer

  1. Charged Point: Use a charged conductor to demonstrate point discharge.

  2. Candle at a Point: Position a candle flame near the charged conductor; the hot gas creates a pathway for charge to flow.

  3. Blown Flame: When the flame is blown away, the electric field can lead to visible discharge from the conductor, showing how point discharge can occur even without direct contact.

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