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How do we determine the electric field due to a continuous charge distribution ? Explain. Electric field due to continous charge distribution |
Answer» SOLUTION :The electric charge is quantized microscopically . The expressions of Coulomb.s Law , superposition principle force and electric field are applicable to only point charges . While dealing with the electric field due to a charged sphere or a charged wire etc. it is very difficult to look at individual charges in these charged bodies . Therefore it is assumed that charge is distributed continuously on the charged bodies and the discrete nature of charges is not considered here. The electric field due to such continuous charge distributions is found by invoking the method of calculus. Consider the FOLLOWING charge object of irregular shape . The entire charged object is divided into a large number of charge elements `Deltaq_(1),Delta_(2),Delta_(3)......Deltaq_(n),......` and each charge element `Delta_(q)` is taken as a point charge . The electric field at a point P due to a charged object is approximately given by the sum of the fields at P due to all such charge elements .  ` <br> `vecE~~(1)/(4piepsilon_(0))((Deltaq_(1))/(r^(2))hatr_(1p)+(Deltaq_(2))/(r^(2))hatr_(2p)+cdots+(Deltaq_(n))/(r_(nP)^(2))hatr_(nP))~~(1)/(4piepsilon_(0))sum_(i-1)^(n)(Deltaqhati)/(r_(iP)^(2))hatr_(iP)`<br> Here `Delta qi` is the `i^(th)` charge element `r_(iP)` is the distance of the point P from the `i^(th)` charge element and `hatr_(iP)` is the unit vector from ith charge element to the point P. However the equation is only an approximation . To incorporate the continuous distribution of charge we take the limit `Deltaq rarr 0 (=dq)` . In this limit the summation in the equation becomes an integration and takes the following form <br> `vecE=(1)/(4piepsilon_(0))int(dq)/(r^(2))hatr` <br> Here r is the distance of the point P from the <klux>INFINITESIMAL</klux> charge dq and `hatr` is the unit vector from dq to point P. Even though the electric fiel for a continuous charge distribution is difficult to evaluate the force experienced by some test charge q in this electric field is still is still given by `vecF=qvecE`. <br> (a) Line charge distribution : If the charge Q is uniformly distributed along the wire of length L, then linear charge density ( charge <klux>PER</klux> unit length ) is `lambda=(Q)/(L)` . Its unit is coulomb per meter `(Cm^(-1))` . The charge present in the infinitesimal length dl is dq `= lambdadl` <br> <img src=) The electric field due to the line of total charge Q is given by `vecE=(1)/(4piepsilon_(0))int(lambdadl)/(r^(2))hatr=(lambda)/(4piepsilon_(0))int(dl)/(r^(2))hatr` (b) Surface charge distribution : If the charge Q is uniformly distributed on a surface of area A then surface density ( charge per unit area ) is `sigma=(Q)/(A)` . Its unit is coulomb per square meter `(Cm^(-2))` .The charge present in the infinitesimal area dA is dq = `sigmaA`. The electric field due to a of total charge Q is given by `vecE=(1)/(4piepsilon_(0)) int(sigmada)/(r^(2))=(1)/(4piepsilon_(0)) sigma int(da)/(r^(2))hatr ` (c ) Volume charge distribution : If the charge Q is uniformly distributed in a volume V then volume charge density ( charge per unit volume ) is given by `rho=(Q)/(V)` . Its unit is coulomb per cubic meter `(Cm^(-3))` The charge present in the infinitesimal volume element dV is dq = `rhodV`. The electric field due to a volume of total charge Q is given by `vecE=(1)/(4piepsilon_(0))int (rhodV)/(r^(2))=(1)/(4piepsilon_(0))rho int (dV)/(r^(2))hatr`. |
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