Question

two metal plates are attached to the two terminals of a 1.5 V battery. how much is required to carry a + 5.0uC charge. (a) from the negative to the positive plate, (b) from the positive to the negative plate

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Work done in moving 5μC charge from negative to positive plate is 7.5 x 10⁻⁶J and work done in moving charge from positive to negative plate is - 7.5 x 10⁻⁶ J .

Explanation:

Given that :

• Potential difference between plates = 1.5 V
• Charge, q = 5μC

To find :

• Work done in moving charge from negative to positive plate
• Work done in moving charge from positive to negative plate

Solution :

• Let, V(a) be the positive plate potential and V(b) be the negative plate potential.
• The potential difference between positive and negative plate is V = V(a) - V(b) = 1.5 V
• The electric potential is the amount of work done in moving a unit positive charge from infinity to that point against the electrostatic forces.
• If a charge, q is moved from potential V1 to V2, then work done, W is

$W =( V2-V1) \times q$

• Now, work done in moving the charge 5μC from negative to positive plate is

$W= (V(a) - V(b))q \\ W = 1.5 \times 5 \times {10}^{ - 6} J \\ W =7.5 \times {10}^{ - 6} J$

• Now, work done in moving the charge 5μC from positive to negative plate is

$W= (V(b) - V(a))q \\ W = - 1.5 \times 5 \times {10}^{ - 6} J \\ W = - 7.5 \times {10}^{ - 6} J$

• Here, negative sign shows that work is done by charge itself in moving from positive to negative plate.