1. 1/1 points | Previous Answers Giancoli6 17.P.006. An electric field of 630 V/
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1. 1/1 points | Previous Answers Giancoli6 17.P.006. An electric field of 630 V/m is desired between two parallel plates 10.0 mm apart. How large a voltage should be applied? 6.3 2. 1/1 points | Previous Answers Giancoli6 17.P.012. What is the speed of a proton whose kinetic energy is 2.9 kev? 7.45e5 m/s 3. 1/1 points Previous Answers Giancoli6 17.P015. A point charge Q creates an electric potential of +148 V at a distance of 29 cm. What is Q? 4.77 nc 4. 1/1 points Previous Answers Giancoli6 17.P.020. An electron starts from rest 70.5 cm from a fixed point charge with Q =-0.200 C. How fast will the electron be moving when it is very far away? 3.0e7 m/s 5. 1/1 points Previous Answers Giancoli6 17.P.032 A 5600 pF capacitor holds plus and minus charges of 20.0 x 10-8 C. What is the voltage across the capacitor? 35.7Explanation / Answer
Sorry we are allowed to solve one question at a time.
but since you want only equations that's why I am giving couple of answer.
Please upload once again for the next questions.
(1) Electric field = potential difference / distance
Potential = 630 *10*10-3 = 6.3 V
(2) Speed of proton
Kinetic energy = (1/2)mV2 where m is mass of proton and V is velocity
Since kinetic energy is given in Kev therefore we have to convert it into J = 2*103*1.6*10-19 J
mass of proton = 1.67*10-23 kg
(3) electric potential = Q/4Pi*eo*r where Q is charge , r is distance
and 4Pi*e0 is constant = 9*109
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