We saw in class that the electric field of a long rod (or cylinder) at a point P

Question

We saw in class that the electric field of a long rod (or cylinder) at a point P located at a distance r bigger than the radius Of the rod is given by E = 2 k lambda/r where the field E is oriented along r, (perpendicular to the rod and away from it), and lambda is the linear charge density (positive in this case). (See figure below). If R is the radius of the rod and r_p is the position of P, (a) write an expression for the potential difference V_P - V_R = -integral_R^P E, dr = -integral_R^r 2 k lambda dr/r = ? (b) If at point P a negative charge -q is released from rest it would be attracted by the rod. Using your answer to part(a) write the expression for the final Kinetic Energy of the charge when it strikes the rod.

Explanation / Answer

a) From the given expression and integrating we get potential difference Vp-Vr = -2klamda[lnr]

considering the range R to r we get

Vp-Vr = -2klamda[ln(R)-ln(r)]

b) Now for a charged particle if we consider the initial kinetic energy as zero.

then at zero potential energy we can write

KE= 1/2mv^2 = -q *potential difference

now from a) the potential difference is -2klamda[ln(R)-ln(r)]

using it in the kinetic energy expression we get,

KE= 1/2mv^2 = -q { -2klamda[ln(R)-ln(r)]} { the charge is negative so minus sign is used for q}

and v is the velocity of the charge

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