Question

A wafer of titanium dioxide $(\kappa=173)$ of area 1.00 $\mathrm{cm}^{2}$ has a thickness of 0.100 $\mathrm{mm}$ . Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a $12.0-\mathrm{V}$ battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part $(\mathrm{b}),$ what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?

          A wafer of titanium dioxide $(\kappa=173)$ of area 1.00 $\mathrm{cm}^{2}$ has a thickness of 0.100 $\mathrm{mm}$ . Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a $12.0-\mathrm{V}$ battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part $(\mathrm{b}),$ what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A wafer of titanium dioxide $(\kappa=173)$ of area 1.00 $\mathrm{cm}^{2}$ has a thickness of 0.100 $\mathrm{mm}$ . Aluminum is evaporated on the parallel faces to form a parallel-plate capacitor. (a) Calculate the capacitance. (b) When the capacitor is charged with a $12.0-\mathrm{V}$ battery, what is the magnitude of charge delivered to each plate? (c) For the situation in part $(\mathrm{b}),$ what are the free and induced surface charge densities? (d) What is the magnitude of the electric field?
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Transcript

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00:01 In this question we have the value of k which is equal to 173, then area a which is equals to 1 .25 centimeter square.
00:11 Next separation between the capacitors plate that is 0 .1mm.
00:18 So let's first find out the capacitancey which is equals to k epsilon node a divided by d.
00:25 Let's substitute the values 173 multiplied by 8 .85 multiplied by 10 to power minus 12 multiplied by 1 .25 multiplied by 10 to power minus 4.
00:39 This whole is divided by 0 .1 multiplied by 10 to power minus 3.
00:47 On a simplification we get the value of capacitor c which is equal to 1 .914.
00:54 1 .914.
00:55 1 .91 .1 .1.
00:56 For nano ferrade.
01:01 Nano ferrade...
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