Two point charges - q - 1 -4mu C and - q - 2 -9mu C are placed 20cm apart- The electric field due to them will be zero on the line joining them a distance of8cm from -q-1-8cm from -q-2-cfrac - 80 - 13 - cm from -q-2-cfrac - 80 - 13 - cm from -q-1-

Question

Toppr Admin · Accepted Answer

Let-apos-s assume electric field is zero at a distance x from first point charge-xA0-Then at the point where electric field will be zero-Ep-0-x21D2-4x2-9-20-x2212-x-2-x21D2-20-x2212-xx-32

Two point charges $q_{1} = 4 μ C$ and $q_{2} = 9 μ C$ are placed $20 c m$ apart. The electric field due to them will be zero on the line joining them at a distance of
$8 c m$ from $q_{1}$
$8 c m$ from $q_{2}$
$\frac{80}{13} c m$ from $q_{2}$
$\frac{80}{13} c m$ from $q_{1}$

Let's assume electric field is zero at a distance $x$ from first point charge.
Then at the point where electric field will be zero:
$E_{p} = 0 \Rightarrow \frac{4}{x^{2}} = \frac{9}{{(20 - x)}^{2}} \Rightarrow \frac{20 - x}{x} = \frac{3}{2}$

Two point charges q1=4μC and q2=9μC are placed 20cm apart. The electric field due to them will be zero on the line joining them at a distance of8cm from q18cm from q28013cm from q28013cm from q1

Let's assume electric field is zero at a distance x from first point charge. Then at the point where electric field will be zero:Ep=0⇒4x2=9(20−x)2⇒20−xx=32

Two point charges $q_{1} = 4 μ C$ and $q_{2} = 9 μ C$ are placed $20 c m$ apart. The electric field due to them will be zero on the line joining them at a distance of
$8 c m$ from $q_{1}$
$8 c m$ from $q_{2}$
$\frac{80}{13} c m$ from $q_{2}$
$\frac{80}{13} c m$ from $q_{1}$

Let's assume electric field is zero at a distance $x$ from first point charge.
Then at the point where electric field will be zero:
$E_{p} = 0 \Rightarrow \frac{4}{x^{2}} = \frac{9}{{(20 - x)}^{2}} \Rightarrow \frac{20 - x}{x} = \frac{3}{2}$