The magnitude of a uniform electric field between the 2 platesis about 2x10^5 N/
ID: 1729457 • Letter: T
Question
The magnitude of a uniform electric field between the 2 platesis about 2x10^5 N/C. If the distance between these plates is 0.2cm,find the potential difference between the plates. Answer in unitsof V. A force of 4.50 x10^-2 N is needed to move a charge of61.0 C a distance of 24.0cm in the direction of a uniformelectric field. What is the potential difference that will providethis force? Answer in units of V. An object with a charge of 10C and a mass of 0.3kgaccelerates from rest to a speed of 11 m/s. Calculate the kineticenergy gained. Answer in units of J. And Through how large apotential difference did the object fall? Answer in units ofV The magnitude of a uniform electric field between the 2 platesis about 2x10^5 N/C. If the distance between these plates is 0.2cm,find the potential difference between the plates. Answer in unitsof V. A force of 4.50 x10^-2 N is needed to move a charge of61.0 C a distance of 24.0cm in the direction of a uniformelectric field. What is the potential difference that will providethis force? Answer in units of V. An object with a charge of 10C and a mass of 0.3kgaccelerates from rest to a speed of 11 m/s. Calculate the kineticenergy gained. Answer in units of J. And Through how large apotential difference did the object fall? Answer in units ofVExplanation / Answer
1. Given: Uniform Electric Field E= 2x105 N/C d= 0.2 cm = 0.002m For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (2x105 N/C)(0.002m) V= 400 N-m/C = 400V (1N/C = 1V/m) 2. Given: Uniform Electric Field F= 4.5x10-2 N q= 61.0 C = 61.0x10-6 C d= 24.0 cm = 0.24m For a uniform electric field, the work accomplished moving aparticle a given distance d is equal to the force required timesthe distance. This, in turn, is equal to the charge of theparticle q, times the electric field, times thedistance: W = Fd = qEd. Fd= qEd F= qE E = F/q E = (4.5x10-2 N)/(61.0x10-6 C) E = 7.38x102 N/C For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (7.38x102 N/C)(0.24m) V= 177 N-m/C = 177V (1N/C = 1V/m) 3. Given: q = 10 C m= 0.3kg v= 11 m/s KE = 1/2mv2 = 1/2(0.3kg)(11 m/s)2 KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V V = (1.65N-m)/(10 C) = 16.5N-m/C = 16.5V (1N/C = 1V/m) V= 400 N-m/C = 400V (1N/C = 1V/m) 2. Given: Uniform Electric Field F= 4.5x10-2 N q= 61.0 C = 61.0x10-6 C d= 24.0 cm = 0.24m For a uniform electric field, the work accomplished moving aparticle a given distance d is equal to the force required timesthe distance. This, in turn, is equal to the charge of theparticle q, times the electric field, times thedistance: W = Fd = qEd. Fd= qEd F= qE E = F/q E = (4.5x10-2 N)/(61.0x10-6 C) E = 7.38x102 N/C For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (7.38x102 N/C)(0.24m) V= 177 N-m/C = 177V (1N/C = 1V/m) 3. Given: q = 10 C m= 0.3kg v= 11 m/s KE = 1/2mv2 = 1/2(0.3kg)(11 m/s)2 KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V V = (1.65N-m)/(10 C) = 16.5N-m/C = 16.5V (1N/C = 1V/m) F= 4.5x10-2 N q= 61.0 C = 61.0x10-6 C d= 24.0 cm = 0.24m For a uniform electric field, the work accomplished moving aparticle a given distance d is equal to the force required timesthe distance. This, in turn, is equal to the charge of theparticle q, times the electric field, times thedistance: W = Fd = qEd. Fd= qEd F= qE E = F/q E = (4.5x10-2 N)/(61.0x10-6 C) E = 7.38x102 N/C For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (7.38x102 N/C)(0.24m) V= 177 N-m/C = 177V (1N/C = 1V/m) 3. Given: q = 10 C m= 0.3kg v= 11 m/s KE = 1/2mv2 = 1/2(0.3kg)(11 m/s)2 KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V V = (1.65N-m)/(10 C) = 16.5N-m/C = 16.5V (1N/C = 1V/m) F= qE E = F/q E = (4.5x10-2 N)/(61.0x10-6 C) E = 7.38x102 N/C For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (7.38x102 N/C)(0.24m) V= 177 N-m/C = 177V (1N/C = 1V/m) 3. Given: q = 10 C m= 0.3kg v= 11 m/s KE = 1/2mv2 = 1/2(0.3kg)(11 m/s)2 KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V V = (1.65N-m)/(10 C) = 16.5N-m/C = 16.5V (1N/C = 1V/m) E = (4.5x10-2 N)/(61.0x10-6 C) E = 7.38x102 N/C For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (7.38x102 N/C)(0.24m) V= 177 N-m/C = 177V (1N/C = 1V/m) 3. Given: q = 10 C m= 0.3kg v= 11 m/s KE = 1/2mv2 = 1/2(0.3kg)(11 m/s)2 KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V V = (1.65N-m)/(10 C) = 16.5N-m/C = 16.5V (1N/C = 1V/m) For a uniform electric field, the potential difference betweenthe plates is equal to the electric field times the distancebetween the plates: V = Ed V= (7.38x102 N/C)(0.24m) V= 177 N-m/C = 177V (1N/C = 1V/m) 3. Given: q = 10 C m= 0.3kg v= 11 m/s KE = 1/2mv2 = 1/2(0.3kg)(11 m/s)2 KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V KE = 1.65kg-m2/s2 = 1.65N-m =1.65J (1J = 1N-m) PE = KE = qV KE/q = V V = (1.65N-m)/(10 C) = 16.5N-m/C = 16.5V (1N/C = 1V/m)Related Questions
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