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aleksandrvk [35]
4 years ago
15

Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac

es is 40 pC/m2, what is the magnitude of the electric potential differences between the two conductors?
Physics
1 answer:
dangina [55]4 years ago
7 0

Explanation:

Relation between electric field and charge density is as follows.

           E = \frac{\sigma}{2 \epsilon}

where,    \sigma = charge density

              \epsilon = permittivity of free space = 8.85 \times 10^{-12}

So,  E_{\text{outside}} = 0

      E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}

or,     E_{inside} = \frac{\sigma}{\epsilon}

Now, formula to calculate the potential difference of two conductors is as follows.

         V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}

It is given that,

           d = 6.0 mm = 6 \times 10^{-3} m

        \sigma = 40 \times 10^{-12} C/m^{2}

Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

        V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}

                     = \frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}      

                     = 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

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The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 c
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3 years ago
A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant
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Answer:

9.96\cdot 10^{-10}J

Explanation:

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k = 3.00 is the dielectric constant

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d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

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U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J

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