33Q.2 all we can say is Ex=20V/cm
(V(x)=-20x+100 so E=-∂V(x)/∂x =20V/cm)
But Ey and Ez remain undetermined with given information... so Enet may be 20V/cm or may be greater tahn 20V/cm but can't be less than 20V/cm
so (b) and (c)
Q1. The electric field and electric potential at a point are E and V respectively.
(a) If E=0, V must be 0
(b) If V=0, E must be 0
(c) If E≠0, V cannot be zero
(d) If V≠0, E cannot be zero.
Q2. The electric potential decreases uniformly from 120V to 80V as one moves on the X-axis from x=-1cm to x=1cm. The electric field in the region is
(a) must be equal to 20V/cm
(b) may be equal to 20V/cm
(c) may be greater than 20V/cm
(d) may be less than 20V/cm
plz giv answers with explanation
-
UP 0 DOWN 0 0 13
13 Answers
33If E=0 then V=constant...(can be zero) (a) is false
also between two equal charges... V>0 but E=0
also if V=0 E is not always zero... consider V=x then E=-1, V=0 at x=0 but E=-1 at x=0 so (b) is false..
potential and field at dipole's perpendicular bisector...
(c) is false... In above example... E=-1 at x=0 but Vat x=0=0
(d) If V=constant... E=0 if V≠0
[12] all false....
kuch mistake hai kya mere expalantion me...??
1for d bt if e=0 n v is work done to bring charge at that pt how v can ve any value cud sum1 explain
331fr 2nd i think ans is can be greter than 20
- ACADBLOCK77321 i always get the value as 20 but how can we prove its greater then 20 ???Upvote·0· Reply ·2019-06-17 08:35:05
33E at that point is zero...
consider..
halfway between two positive charge.. V>0 and E=0
E=0 at that point... not in surrounding.... so in bringing charge work is done against E in surrounding....
11st one's answer shd b option c...a and b are wrong itz in the board bookks...fr rest 2 options reach the coclusion by v=e.d
33(c) is also wrong...
Consider. halfway between dipole... V=0 but E is not zero...
106thx priyam for both questions..
yes first one is (none) and second one is (b) (c)