Two 2.1cm diameter disks face each other, 2.9mm apart. They are charged to ±10nC. a) what is the electric field strength between two disks? b) a proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

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Two 2.1cm diameter disks face each other, 2.9mm apart. They are charged to ±10nC.   a) what is the electric field strength between two disks?   b) a proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?

Annie Gonzalez · Accepted Answer

Step 1 Given: Diameter of disk =2.1cm Radius of disk =1.05cm =1.05×10−2m Distance between two disks, s=2.9mm =2.9×10−3m Charge on disks =±10nC =±10×10−9C Permitivity of free space, ξ0=8.854×10−12Fm Step 2 (a) Electric field strength can be calculated as E=σξ0 Where, σ is the surface charge density σ=QA Where A is the area A=πr2 Therefore, E=Qξ0πr2 =10×10−9C8.854×10−12×(1.05×10−2)−2 =3.26×106NC Step 3 Electric force =qE Also, Force=ma Since, both the force experienced by the same charge particle Therefore, qE=ma a=qEm where m is the mass of the proton m=1.67×10−27kg and e is the charge of the proton a=1.6×10−19×3.26×1061.673×10−27 a=3.12×1014ms2 Step 4 Using the equation of motion s=ut+0.5at2 initial velocity of the particle, u=0 s=0.5at2 t=2sa =2×2.9×10−33.12×1014 =4.3×10−9s Step 5 (b) Speed of the proton Again using the equation of motion v=u+at v=0+3.12×1014×4.3×10−9 =13.42×105ms Step 6 Answer; (a) Electric field strength between two disks =3.26×106NC (b) launch speed of the proton to just barely reach the positive disk =13.42×105ms

Karen Robbins · Accepted Answer

Answer:   E=2.88⋅106NC   v=0.742⋅106ms   Explanation:   The two disks form a parallel-plate capacitor, which will cause an electric field equal to:  E=Qe0A  Where Q is the charge of the disks, A is the area of the disks and e0 is the vacuum permisivity equal to 8.85⋅10−12C2Nm2:  E=8⋅10−9C(14π(0.02m)2)⋅8.85⋅10−12C2Nm2=2.88⋅106NC  Now, for the second part of the problem, we can use conservation of energy. The addition of potential and kinetic energy at launch point should be equal to the addition at the positive disk. Because the proton has positive charge, the potential energy of the proton will increase as its distance to the negative disk increases too. This is because the proton will be attracted towards the negative disk. The potential energy is given by:  Ep=V⋅q  Where V is the difference in potential (voltage) between the disks. In a parallel-plate capacitor:  V=E⋅d, where d is the difference in position with the frame of reference. Our frame of reference will be the negative disk.   q is the charge of the proton.   The kinetic energy is given by:  Ek=12mv2  Then:  Ep1+Ek1=Ep2+Ek2  V1⋅q+12mv12=V2⋅q+12mv22∣v2=0,Ep1=0  12mv12=E⋅d⋅q  v=2Edqm=2⋅2.88⋅106Nm⋅0.001m⋅1.6⋅10−19C1.67⋅10−27=0.742⋅106ms

alenahelenash · Accepted Answer

(A) electric field between to charged disk =E→=QAξo Q=charge on each disk=9.0nC r=disk radius=1.5e−02m A=surface area of each disk=πr2=0.0007069m2 ξ0=8.854×10−12C2/N m2 ⇒E→=QAξo=1.438e+06NC (B) kinetic energy required by the proton to move from negative disk to positive disk =K K=q△V=qEd q=1.602e−19C E=1.438e+06NC d=gap between disk=1.0mm ⇒K=2.304e−016J if the launch speed =v K=mv22; mass of proton =m=1.673e−027kg v=2Km=5.249e+05ms

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