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Feel the Potential of Physics Answers 1. A charge of 5.0 μC is moved from one location to another in an electric field. 25.0 J of work is done. Find the potential difference. W V = q 25.0 J = 5.0 x 10-6 C V = 5.0 x 106 J/C = 5.0 x 106 V 2. When an ion accelerates through a potential difference of 2850 V, its electric potential decreases by 1.378 x 10-15 J. What is the charge on the ion? W V = q = = q W V 1.378 x 10-15 J 2850 V q = 4.8 x 10-19 C qV = W q = Energy V 3. If an electron is accelerated through a potential difference of 100 V, what is the final speed of the electron? W V = q Work = Energy KE = ½ m v2 W = Energy = q V = ( 1.6021 x 10-19 C )( 100 V ) = 1.6021 x 10-17 J 3. If an electron is accelerated through a potential difference of 100 V, what is the final speed of the electron? Energy = KE = 1.6021 x 10-17 J KE = 1/2 mv2 v = 2 KE = m v = 5.93 x 106 m/s 2 ( 1.6021 x 10-17 J ) 9.11 x 10-31 kg 4. A capacitor hooked up to a 6.0-V battery acquires 8.0 μC of charge. (a) What is the capacitance of the capacitor? Q = CV C = Q V C = = = = = 8.0 x 10-6 C 6.0 V 1.3 x 10-6 C/V 1.3 x 10-6 farads 1.3 x 10-6 F 1.3 μF 4. A capacitor hooked up to a 6.0-V battery acquires 8.0 μC of charge. (b) If the plates are 0.50 mm apart, what is the electric field intensity between the plates? milli- V E = 6.0 V = d 0.00050 m E = 12 000 V/m = 1.2 x 104 V/m ( or 0.50 x 10-3 m ) 5. An industrial capacitor can have a capacitance of 2.50 F. Its plates are 0.75 mm apart. If this capacitor is connected to a potential difference of 120 V, find (a) the electric field intensity between the plates milli- V E = 120 V = d 0.00075 m E = 160 000 V/m = 1.6 x 105 V/m ( or 0.75 x 10-3 m ) 5. An industrial capacitor can have a capacitance of 2.50 F. Its plates are 0.75 mm apart. If this capacitor is connected to a potential difference of 120 V, find (b) the charge stored on the capacitor Q = CV = ( 2.50 F )( 120 V ) Q = 300 C