Assume a uniform electric field everywhere, directed toward to the right, x-dire

Question

Assume a uniform electric field everywhere, directed toward to the right, x-direction (see figure). The magnitude of the field is E=500N/C. The (x,y) coordinates of the locations are: A(0m,0.30m), B(0m,0m), C(0.40m,0.20m). Let's calibrate the potential V at B zero that V(0,0)=0 (a) Which statement is true? (S1) V at all locations (A,B,C) is the same because E is uniform. (S2) V at location A and at B is the same and V at C is lower than V at A. (S3) V at location A and at B is the same and V at C is higher than V at A. (S4) V at A is highest, V at B is lowest.(b) calculate V from E: (1) write general formula to get V(r_2) from vector E when V(r_1)=0 is given. (2) find V(A) and V(C) with values in V (volt) (3) when is V higher? towards the positive or towards the negative charges that are responsible for E? Q11 (1P): A proton is accelerated through a potential difference of 100V. What is the change in potential energy of the proton? Express the energy in J and eV.

Explanation / Answer

a) S3 statement is true that is V=E.dr here E=500i,dr of A=0i+.3j,B=oi+0j,C=0.4i+0.2j then

VA=0,VB=0,VC=500*0.4=200V there fore Va=Vb=0,Vc higher than Va

b) 1)potential at A is Vr2=E.dr2 hereE=500i,dr2=0i+.3j

2)

)potential at A is Vr2=E.dr2 hereE=500i,dr2=0i+3j V=(500i).(oi+0.3j)=0

3)V is higher towards positivedirection

Q11)change in otencial energy dw=qdV

here q=1.6*10-19c,dV=100v

then dW=1.6*10-19*100=1.6*10-17j

dW=100*1.6*10-19=100eV

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