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meriva
3 years ago
15

A battery of emf 5V and internal resistance 2ohm is joined to a resistor of 8ohm.Calculate the terminal potential difference. ​

Physics
1 answer:
nata0808 [166]3 years ago
3 0

Answer:

4V

Explanation:

First, we calculate the total resistance to the given battery cell of emf 5V. The total resistance is the sum of all the resistance in the cell i.e.

Total resistance = 2Ω + 8Ω = 10Ω

Using ohms law equation to calculate the current passing through the battery cell:

V = IR

Where; V = voltage, I = current, R = resistance

5V = I × 10Ω

I = 5/10

I = 0.5A

Terminal voltage is calculated by the us of the following equation:

V=emf−IR

Where; R is internal resistance

V = 5 - (0.5 × 2)

V = 5 - 1

V = 4V

Therefore, the potential difference across the terminals of the battery cell is 4V

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A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

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The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

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=>     r =  \frac{0.12}{2}

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Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

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3 years ago
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