Line Integral of Electric Field
Description:
.fraction {
display: inline-block;
vertical-align: middle;
margin: 0 0.2em 0.4ex;
text-align: center;
}
.fraction > span {
display: block;
padding-top: 0.15em;
}
.fraction span.fdn {border-top: thin solid black;}
.fraction span.bar {display: none;}
.intsuma {
position: relative;
display: inline-block;
vertical-align: middle;
text-align: center;
}
.intsuma > span {
display: block;
font-size: 70%;
}
.intsuma .lim-up {
margin-bottom: -1ex;
}
.intsuma .lim {
margin-top: -0.5ex;
}
.intsuma .sum {
font-size: 1.5em;
font-weight: lighter;
}
.intsuma .sum-frac {
font-size: 1.5em;
font-weight: 100;
}
.sy {
position: relative;
text-align: center;
}
.oncapital, .onsmall {
position: absolute;
top: -1.2em;
left: 0px;
width: 100%;
font-size: 70%;
text-align: center;
}
.onsmall {
top: -0.7em;
}
The multiplication of electric field and the path length gives Line integral of electric Field.
Let there be two point A and B on a non uniform electric field (magnitude and direction at every point is different). The irregular line denotes the path travelled by charge. With changing Electric field, the path travelled by a charge also changes at every point on the curve.
To find line integral at every point, take a very small length dl on the path. So that dl makes a certain angle θ with the electric field. On resolving dl into two components as dlcosθ and dlsinθ the component
of dl along E is dl cosθ. As shown in figure.
Potential difference dV = Edr.
So line integral to move charge from A to B = A∫B EdrCosθ
VA - VB = A∫B E→.dr→
To move charge from B to A (VA - VB) = - B∫A E→.dr→
Note−
in a variable electric field, potential difference between two point is equal to negative of line integral of electric field.