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A positive point charge q+2pC of mass m-200g is released from rest in a uniform electric field E-2x10 N/C directed along the s-axis as shown in figure, the distances between tow plate is 20cm, answer Q, Q20, and Q21. Q15) The acceleration of charge in m/s² is: a) 10 b) 20 c) 30 d) 40 Q16) The speed of the point charge when it reached to the negative plate in m/s is: a) 1.8 b) 2 c) 2.3 d) 2.8 Q17) The kinetic energy of charge when it reached to the negative plate in joul is: a) 0.4 b) 0.6 c) 0.8 d) 1.6 13 00000

A positive point charge q+2pC of mass m-200g is released from rest in a uniform electric field E-2x10 N/C directed along the s-axis as shown in figure, the distances between tow plate is 20cm, answer Q, Q20, and Q21. Q15) The acceleration of charge in m/s² is: a) 10 b) 20 c) 30 d) 40 Q16) The speed of the point charge when it reached to the negative plate in m/s is: a) 1.8 b) 2 c) 2.3 d) 2.8 Q17) The kinetic energy of charge when it reached to the negative plate in joul is: a) 0.4 b) 0.6 c) 0.8 d) 1.6 13 00000

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Publisher:Raymond A. Serway, Chris Vuille
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A positive point charge q+2pC of mass m-200g is released from rest in a uniform electric field E-2 x 10 N/C directed along the s-axis as shown in figure, the distances between tow plate is 20cm, answer Q, Q20, and Q21. Q15) The acceleration of charge in m/s2 is: a) 10 b) 20 c) 30 d) 40 Q16) The speed of the point charge when it reached to the negative plate in m/s is: a) 1.8 b) 2 c) 2.3 d) 2.8 Q17) The kinetic energy of charge when it reached to the negative plate in joul is: a) 0.4 b) 0.6 c) 0.8 d) 1.6 a) 900 volt Q18) Two point charge q₁-3 ne at 3cm from charge q6 ne,then electric potential at point 3cm from q, is: b) 1800 volt c)-900 volt d)-1800 volt e) 600 volt Q19) In order for two object to have same temperature they must: a) be in thermal contact together b) have the same hotness or coldness when touched c) be in thermal equilibrium d) have all of the properties above e) have only a and b properties c) 123 + d) 246.4 + Q20) The change in kinetic energy (in kJ) for a 2000 kg car moving on a horizontal road to increase its velocity from (2î+3)) m/s to (5î+12)) m/s is: a)156 b) 151 Q21) Total mechanical energy of any system is : a) The sum of the kinetic energy and the work b)Always the work done by gravity c) The sum of potential energy and work d) The sum of the kinetic and potential energy at any point |||||||||| e) 114.4
Transcribed Image Text:A positive point charge q+2pC of mass m-200g is released from rest in a uniform electric field E-2 x 10 N/C directed along the s-axis as shown in figure, the distances between tow plate is 20cm, answer Q, Q20, and Q21. Q15) The acceleration of charge in m/s2 is: a) 10 b) 20 c) 30 d) 40 Q16) The speed of the point charge when it reached to the negative plate in m/s is: a) 1.8 b) 2 c) 2.3 d) 2.8 Q17) The kinetic energy of charge when it reached to the negative plate in joul is: a) 0.4 b) 0.6 c) 0.8 d) 1.6 a) 900 volt Q18) Two point charge q₁-3 ne at 3cm from charge q6 ne,then electric potential at point 3cm from q, is: b) 1800 volt c)-900 volt d)-1800 volt e) 600 volt Q19) In order for two object to have same temperature they must: a) be in thermal contact together b) have the same hotness or coldness when touched c) be in thermal equilibrium d) have all of the properties above e) have only a and b properties c) 123 + d) 246.4 + Q20) The change in kinetic energy (in kJ) for a 2000 kg car moving on a horizontal road to increase its velocity from (2î+3)) m/s to (5î+12)) m/s is: a)156 b) 151 Q21) Total mechanical energy of any system is : a) The sum of the kinetic energy and the work b)Always the work done by gravity c) The sum of potential energy and work d) The sum of the kinetic and potential energy at any point |||||||||| e) 114.4
Expert Solution
Step 1

Charge on particle = 2×10^-6 C.

Mass = 200 gm = 200 × 10^-3 kg = 0.2 Kg 

Electric field = 2 × 10^6 N/C. 

Force on charge = E × Q = (2×10^-6) × ( 2 × 10^6) 

Force = 4N.( In the direction of electric field). 

Acceleration= (force /mass ) =( 4/0.2)= 20m/s^2. 

Distance between plate = 20cm = 0.20 m 

Use the relation:- 

V^2 = U^2 + 2× a × s 

U = initial velocity = 0m/s 

a = acceleration = 20m/s^2 

S = distance = 0.20m 

Putting the value :- 

V^2 = (2× 20 × 0.20 ) = 8 

Hence , V = 8^0.5 = 2.828 m/s^2 

 

 

 

 

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