Using Gauss theorem, the electric field distribution around a straight line can be found as follows:
E=λ/(2π*εo*r)
According to the fact that the work done by electric field force has nothing to do with the path, the movement of charge Q from point A to point B can be regarded as the movement along the electric field line from a straight line to a straight line.
Therefore, the upper integral limit W=∫E*dr of electric field force work is b, and the lower integral limit is a (assuming a.
This integral can be obtained by the following formula:
W=λ/(2π*εo)*ln(b/a)
2. In vacuum, there are +q and -3q on two concentric spheres with radii r and 2R.
If a charged particle with a +Q charge is released at rest in the inner sphere, the kinetic energy of the particle when it reaches the outer sphere is ().
We know that in the state of electrostatic equilibrium, the charge can only be distributed on the inner surface of charged body. In this way, the outer surface of the inner spherical shell is charged with +q, so the inner surface of the outer spherical shell is bound to be charged with -q, while the outer surface of the outer spherical shell is charged with -2q.
Let the potential between the inner and outer spherical shells be u.
According to U=∫E*dr, the upper limit of integral is 2R and the lower limit is r.
e = q/(4 π * ε o * r 2)
Find u = q/(4 π * ε o) * [1/r-1/(2r)].
=q/[(4π*εo)*2R]
Thus, according to the kinetic energy theorem, it can be found that when the charge of Q reaches the outer sphere, the velocity is V and the kinetic energy is Ek.
Ek= 1/2*m*V^2=U*Q
Solve Ek=Uq/[(4π*εo)*2R]
3. the key is to find the potential difference u from k to 2mm.
According to the formula of potential difference between cylinders, there are:
300=k*ln(R/r)=k*ln9
u = k * ln[(r+2mm)/r]= k * ln(2.5/0.5)= k * ln5
Where k represents a constant.
So you can find you.
According to the kinetic energy theorem, let the velocity when the distance k is 2mm be V 1.
There are obviously:
1/2*m*V 1^2=U*e
You can find V 1.
2) The speed to reach a is V2.
Yes: 300 * e = 1/2 * m * v2 2.
Where e =1.6 *10 (-19) c and m = 9.1*10 (-31) kg.
4. The capacitance formula of1) capacitor is c = ε s/d.
When the copper plate is inserted into the capacitor, the field strength in the copper plate is 0, which is equivalent to changing the distance between the two plates from D to d-d'
So the capacitance at this time is: C'=εo*S/(d-d') (please forgive me for not remembering the formula of capacitance clearly).
Regardless of the distance between the copper plate and the polar plate.
2) The charging q of the capacitor after charging is:
Q=C'*Vo(Vo stands for potential difference)
The electric field energy of the capacitor is w 1 = Q2/(2c').
After removing the copper plate, the capacitor remains charged, but the capacitance changes.
At this time, the electric field energy is W2.
W2=Q^2/(2C)
According to the conservation of energy, external work is equal to the change of electric field energy, and let external work be w.
Obviously:
W=W2-W 1
2R's ball is -3q, not -2q. And why E = q/(4π * ε o * R 2) is clear.
Originally, the spherical shell of 2R was charged with -3q, but the spherical shell of R was charged with +q, resulting in the inner surface of the spherical shell of 2R being charged with -q (thus making the ball of 2R uncharged).
According to the conservation of charge, it can be known that the spherical shell of 2R can only be charged to -2q (the total charge is -3q=-q+-2q).
As for why, e = q/(4 π * ε o * r 2)? This is based on Gauss theorem.
∫∫E*dS=q/εo