Electric charge is distributed uniformly along a thin rod of length a, with total charge Q . Tak

Electric charge is distributed uniformly along a thin rod of...

Key Concepts

Electric Potential

This is a measure of the work required per unit charge to bring a test charge from a reference point, commonly chosen at infinity, to a specific location in an electric field. In problems involving continuous charge distributions, the electric potential is obtained by integrating the contributions from infinitesimally small elements of charge, with the potential of each element determined by its charge and distance from the point of interest.

Continuous Charge Distribution Integration

When dealing with objects such as a uniformly charged rod, the total electric potential is calculated by integrating the contributions from each infinitesimal element of the charge distribution. This requires setting up an integral where each element has a value proportional to an established charge density, and summing the resulting potentials determined by their respective distances from the observation point.

Uniform Charge Distribution

A uniform charge distribution means that the charge is spread evenly over a given length, area, or volume. For a thin rod, this implies a constant linear charge density, which simplifies the integration process used to calculate the potential by ensuring that every small segment of the rod contributes equally per unit length.

Far-Field Approximation

When the observation point is at a distance much larger than the dimensions of the charge distribution, the potential approaches that of a point charge. This approximation simplifies calculations by allowing the extended charge distribution to be treated as if all the charge were concentrated at a single point, providing a consistency check for the computed potential in the limit of large distances.

Reference Potential

The choice of zero potential at infinity is a common convention in electrostatics. It sets the baseline against which all other potentials are measured, ensuring that the computed electric potential values are relative to a fixed, consistent standard.

Transcript

00:01 In the first part we want to find the electric potential at point p which is at a distance of x from the right end of the road so here first we will divide the road into infinitesimal portions so here we will consider an infinitesimal portion that is having length of dr and having charge on this portion is dq and this small portion is at a distance of r from the right end of the road and the value of this r goes from 0 to a when we move from right to the left of this road now first we will write the relation for electric potential due to this small portion at point p so we have dvp is equal to 1 upon 4 pi afts a not the charge on this small portion divided by by the distance of this small portion from point p which is x plus r as we are given in the problem that the charge is distributed uniformly on the road so we can find the charge on the small portion by using the relation dq equal to the charge per unit length that is total charge divided by total length of road into length of this portion which is dr so by putting the value of dq we will get the potential at point p due to this small portion equal to 1 upon 4 pi epsilon not into q dr divided by a whole divided by x plus r by rearranging we can write this as a electric potential due to the small portion at point p equal to 1 upon 4 pi epsilon not times a into qdr upon x plus r now this is the erratic potential at point p due to a small portion under what now we can find the total erratic potential due to this rod at point p by integrating this relation that is erratic potential at p due to the rod equal to integration of dvp from 0 to a this gives us integration from 0 to a 1 upon 4 pi epsilon not a q d r divided by x plus r by by writing the constant terms in front of integral sign we will have vp equal to here charge on the rod is constant so q upon 4 pi epsilon not and length of rod is also constant into integral from 0 to a 1 upon x plus r d r now by using the relation integral from integral of d x divided by a x plus b which is equal to 1 upon a log of a x plus b we can write that integral as vp equal to q divided by 4 pi a a a of x plus r here the coefficient of x is 1 and here limits are from 0 to a we can put the limit shear q upon 4 pi a a a plus n of a plus r and of sorry l enough x plus a here we are integrating with respect to r so we will put the value of r here minus l and of x plus 0 by using the formulas for logarithm we can write q upon 4 pi a a epsilon not times a l and of x plus a divided by x so this is the electric potential due to this rot at point p now in the b part of this problem again we have the rod the length of this road is a and we want to find directly potential at point p which is directly above the right end of the road and this point is named as r let this point is at a distance y from the end here again we will assume a small portion on the rod at a distance of r from the right end let the length of the small portion is dr and it has a charge of dq now the distance of this small portion from point r can be calculated by using phatagros serum and that will be equal to r squared plus x squared hold to the power half again we can write here the contribution of electric potential due to this small portion which is dvr is equal to 1 upon 4 pi a epsilon not times the charge on the small portion divided by the distance of small portion from point r which is which is which is which is y squared which is r squared plus y secured sorry i did a little mistake here in the diagram here the distance of the small portion from point r is r squared plus y secured again we can write the charge on the small portion equal to the total charge on the rod divided by a length of the road that is the joy density which is defined as charge per unit length into length of the small portion so we have the contribution of electric potential due to the small portion at point r equal to 1 upon 4 pi epsilon not and we have dq equal to qdr divided by a upon r squared plus y squared by rearranging we can write this as dvr is equal to 1 upon 4 pi epsena not times a qdr divided by r squared plus y squared now the total electric potential due to this rot can be computed by integrating the above value that is electric potential due to rot at point r is equal to integration of dvr from 0 to a and we have dvr equal to 1 upon 4 pi epsilon not a qdr divided by square root of r squared plus y squared we can write vr as q divided by 4 pi epsilon not a and integration from 0 to a 1 upon y squared plus r squared whole half into d r by using mathematics we can write v are equal to q divided by 4 pi epsilon not times a and the integration of 1 divided by square root of y square plus r squared that is equal to log of square root of y squared plus r squared alien of square root of y square plus r square plus r and here we have limits from zero to a here we have used the integral that is one upon square root of a square plus x square d x is equal to log of x plus a square plus x sugar now here we will put the limits first we will put r equal to a then we will put r equal to zero so this uses aetic protection at point r is equal to q divided by 4 pi a xena naught a and we have an of of y sugar plus a sugar plus a minus ln of y sugar plus 0 plus 0 this gives us we are is equal to q divided by 4 pi xxna n times a and here we have ln of y sugar plus a sugar whole square root plus a minus and of y now we can write this as we are that is electric protection at point r is equal to two divided by four pi a epsilon not a and by using the formulas for the logarithm we can write this as y -sucker plus a square plus a plus a divided by y so this is the relation for electric protection at point a due to this rot now in the next part of this problem here we will assume that the distance x that is the distance of point p from the road is very very greater than as compared to the length of rot as we have already calculated in part a of this problem that electric potential due to rot at point p is equal to q upon 4 pi epsilon not a log of x plus a divided by x we can write this as q upon 4 pi epsilon not times a log of 1 plus a upon x now by using the expansion of 1 plus x that is equal to 1 plus u that is equal to u minus u square divided by 2 and the next higher added terms by using this expansion we can write this as potential at point p equal to q divided by 4 pi a epsilon not a into expansion of log here is u here u is a upon x so we can write a upon x minus half into a upon x secure and we have next terms of a upon x whole cube.

18:03 Then we will have a upon x to the power for neglecting all these higher added terms...

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