A-3.18-hc point charge and a 6.36-c point charge are a distance l apart. where should a third point charge be placed so that the electric force on that third charge is zero? (hint: solve this problem by first placing the -3.18-hc point charge at the origin and place the 6.36-pc point charge at x =-l.)

Explanation:

Let x be the distance of - 3.18 μC from the said third charge.Then the distance of 2 nd charge will be L + x . It is so because the point of zero field will not be between the charge because of opposite nature of charges. It will be on the other side of smaller charge.Applying formula of electric field we have

k x3.18 x 10⁻⁶ / x²  = k x- 6.36 x  10⁻⁶ /( L+ x)²

=( L+x)²/x² = 6.36/3.18=2

(L + x) /x = √2 = 1.414

L + x = 1.414 x

.414 x = L

x = L/.414

x = 2.4 L

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .

ammonium chloride & tallow