The potential energy of system of two equal negative point charges of $2 μ C$ each held 1 m apart in air is ( $k = 9 \times 10^{9} S I u n i t$ )
$36 J$
$3.6 \times 10^{- 3} J$
$3.6 \times 10^{- 2} J$
$3.6 J$

Question

The potential energy of system of two equal negative point charges of 2mu C each held 1 m apart in air is -k - 9 times 10-9- SI -unit- 36 -J3-6 times 10-3- J3-6 times 10-2- J3-6 -J

Toppr Admin · Accepted Answer

The potential energy of system isU-kq1q2rU-kq2r -xA0- -xA0- -q1-q2-q-or U-9-xD7-109-xD7-2-xD7-10-x2212-6-21or U-9-xD7-109-xD7-4-xD7-10-x2212-12or U-3-6-xD7-10-x2212-2J

The potential energy of system of two equal negative point charges of 2μC each held 1 m apart in air is (k=9×109SIunit) 36J3.6×10−3J3.6×10−2J3.6J

The potential energy of system isU=kq1q2rU=kq2r [q1=q2=q]or U=9×109×(2×10−6)21or U=9×109×4×10−12or U=3.6×10−2J

The potential energy of system of two equal negative point charges of $2 μ C$ each held 1 m apart in air is ( $k = 9 \times 10^{9} S I u n i t$ )
$36 J$
$3.6 \times 10^{- 3} J$
$3.6 \times 10^{- 2} J$
$3.6 J$

The potential energy of system is
$U = k \frac{q_{1} q_{2}}{r}$
$U = k \frac{q^{2}}{r}$ $[q_{1} = q_{2} = q]$
or $U = 9 \times 10^{9} \times \frac{(2 \times 10^{- 6})^{2}}{1}$
or $U = 9 \times 10^{9} \times 4 \times 10^{- 12}$
or $U = 3.6 \times 10^{- 2} J$