IIT JEE , AIEEE Forum - Physics , Electricity

city_gal1234

1)a certain region of space is to be protected from external field.suggest 2 methods.

2)two point charges Aand B of unkown magnitude and sign are placed d distance apart.the electric field is zero at a point not bet the charges but on line joining them and closer to A.write two essential conditions for this to happen

joyfrancis

Let charge of magnitude q1 be placed of A and q2 on B. The condition will never be possibile when the charges are of the same sign .

.: either (i)q1>0 , q2<0 or (ii)q1<0 and q2>0

In (i)

If q1 is positive it will apply a force in the neg dire of x on the test charge and q2 being negative will apply a force in the + x dir. Now since distance of q1 from the test chatge is less , the forces on the test charge would be balanced only if |q1|<|q2|

In (ii) also because of the same reason |q1|<|q2|.

So the conditions possible are

q1<0,q2>0,|q1|<|q2|

& , q1>0, q2<0, |q1|<|q2|

joyfrancis

For the first part , you can create a shell around the given reigin , since electric field in a shell is always 0

neeraj_agarwal_1990

1)surround the region by a conducting shell of any shape that u want .....any metallic chell acts as a field blocker...

spideyunlimited

for 1 ) ... use spherical shell , as all charge collects on outer surface thus no electric field is present inside the shell

for 2 ) at point at distance r from one of the charges (A), electric field is 0.

so kQb/ (r+d)sq + kQa/ (r)sq. = 0
for sum to be zero, first of all Qa and Qb must have opp. sign else they wont cancel.
secondly, denominator of Qa ie. r square is smaller than that of Qb which is (r+d) square.
so to make the two terms of equal magnitude , mag. of Qa must be less than mag. of Qb.

so the two conditions are
1) Qa.< 0, Qb > 0
2) | Qa| < | Qb |

PLZ RATE:)

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