A 1.0cm * 1.0cm parallel-plate capacitor has a 2.0mm spacing. The electric field
ID: 2110243 • Letter: A
Question
A 1.0cm * 1.0cm parallel-plate capacitor has a 2.0mm spacing. The electric field strength inside the capacitor is 1.3*10^5 V/m
A) What is the potential difference across the capacitor? DeltaV=______V
B) How much charge is on each plate? q1,q2=_____C
2) A 20nC charge is moved from a point whereV=150V to a point where V=-90V
A) How much work is done by the force that moves the charge? W=_____
3) An electron has been accelerated from rest through a potential difference of 1000V.
A) What is the electron's speed? v=_____m/s
Explanation / Answer
The E-field inside a PP capacitor is given by ;
E = D/e(o)
Where "D" is the surface charge density on either plate (they are equal in magnitude)
E = Q/Ae(o)
Q = e(o)EA
Where "A" is the plate area in m^2 and e(o) is the dielectric constant in vacum.
2)Q will have same magnitude on either plate (electrode), but of opposite sign.
If the potential at B is 150V , then it should remain 150V whether you move a 20nc charge or 100nc charge. Remember electric potential at a point due to a combination of charges depends only on the charges creating that potential. Potential energyhowever will depend on whether they system of charges contain 20nc or whether it contains 10nc.
3)Ek = mv^2 / 2
v = ((2xEk) /m)^0.5
v = ((2)(1.602x10^-16))/(9.109x10^-31))^0.5
v = 18 754 733.73 m/s
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