pratham ashish
Last Activity: 15 Years ago
hi,
it's very easy just notice that E is vector but V is scalar.
just mark q at both -a & +a , u'll will see (mathematically/physically) that ,at the points nearer to RHS of -a the distance tends to zero hence E tends to ∞( it 's +∞ bcz distace from -a will b + if we r on RHS of a point)
at the mid (at 0) E wud b zero as E of both cancels each other
and LHS of +a it wud -∞(near to +a chrge at this pt wud dominate over the othr as r tends to zero E due to tis charge wud tends to ∞ & distance from this pt to any point on LHS wud b - )
similarly on LHS of -a
r(distace) is -ve &just on LHS of -a E wud tends to-∞ as r tends to 0 & E wud tend to 0 at x= -∞
onRHS of +a
first E ->+∞ just on RHS & on going further it wud decrease & -> 0 at x= +∞
the graph wud b continuous LHS of -a, b/w -a &a and RHS of a
it wud b discont. at -a &a only .
so will b the V graph vs x (u can use the relation dV/Dx= -E(x) )