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mr_godi [17]
2 years ago
6

Two large parallel plates are 2.0 cm apart. The strength of the electric field between them is 50.0 N/C . What is the difference

in voltage between the plates?+10 V+1.0 V−1.0 V0 V
Physics
1 answer:
diamong [38]2 years ago
6 0

Express the distance in meters:

d= 2 cm =0.02 m

V = Ed = 50 N/C * 0.02 m = 1 N/C m = 1 V = +1.0 V

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You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
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When they are connected in series

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From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

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Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

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This means that the 50 W bulb glows more than the 100 \ W bulb

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4 years ago
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