Electric potential at the mid point of two identical point charges placed at some distance is V . If 25 % of charge of one point charge is transferred to another then new potential at mid point will be -

Question

Electric potential at the mid point of two identical point charges placed at some distance is V . If 25 % of charge of one point charge is transferred to another then new potential at mid point will be - (A) (15 V/16) (B) V (C) (3 V/4) (D) (5 V/4)

Tardigrade Admin · Accepted Answer

V1=(K Q/r)× 2=(2 K Q/r) Solution

Q-(Q/4)Q+(Q/4) =(3 Q/4)(5 Q/4) V2=(K ((3 Q/4))/r)+(K ((5 Q/4))/r) =(K Q/r)((3/4) + (5/4))=(k Q/r)((8/4))=(2 K Q/r)=V

Q. Electric potential at the mid point of two identical point charges placed at some distance is $V$ . If $25%$ of charge of one point charge is transferred to another then new potential at mid point will be -

$\frac{15 V}{16}$

$V$

$\frac{3 V}{4}$

$\frac{5 V}{4}$

$V_{1} = \frac{K Q}{r} \times 2 = \frac{2 K Q}{r}$

$Q - \frac{Q}{4} Q + \frac{Q}{4}$
$= \frac{3 Q}{4} \frac{5 Q}{4}$
$V_{2} = \frac{K ( \frac{3 Q}{4} )}{r} + \frac{K ( \frac{5 Q}{4} )}{r}$
$= \frac{K Q}{r} (\frac{3}{4} + \frac{5}{4}) = \frac{k Q}{r} (\frac{8}{4}) = \frac{2 K Q}{r} = V$

Q. Electric potential at the mid point of two identical point charges placed at some distance is V . If 25% of charge of one point charge is transferred to another then new potential at mid point will be -

1615V​

V

43V​

45V​

V1​=rKQ​×2=r2KQ​ Q−4Q​Q+4Q​ =43Q​45Q​ V2​=rK(43Q​)​+rK(45Q​)​ =rKQ​(43​+45​)=rkQ​(48​)=r2KQ​=V

Q. Electric potential at the mid point of two identical point charges placed at some distance is $V$ . If $25%$ of charge of one point charge is transferred to another then new potential at mid point will be -

$\frac{15 V}{16}$

$V$

$\frac{3 V}{4}$

$\frac{5 V}{4}$

$V_{1} = \frac{K Q}{r} \times 2 = \frac{2 K Q}{r}$

$Q - \frac{Q}{4} Q + \frac{Q}{4}$
$= \frac{3 Q}{4} \frac{5 Q}{4}$
$V_{2} = \frac{K ( \frac{3 Q}{4} )}{r} + \frac{K ( \frac{5 Q}{4} )}{r}$
$= \frac{K Q}{r} (\frac{3}{4} + \frac{5}{4}) = \frac{k Q}{r} (\frac{8}{4}) = \frac{2 K Q}{r} = V$