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1 month ago

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A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

hat is the kinetic energy of this charge at x = 4m?

1 answer:

ValentinkaMS [3.4K]1 month ago

8 0

The force due to electricity on the charge is calculated by multiplying the charge by the intensity of the electric field:
$F=qE$
in our scenario, where $q=3 \mu C= 3 \cdot 10^{-6} C$ and $E=2 kN/C=2000 N/C$ , resulting in the force of
$F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N$

Initially, the kinetic energy of the particle is at zero (as it remains stationary), which means its final kinetic energy is equal to the work performed by the electric force over a distance of x=4 m:
$K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J$

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Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

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$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

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V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² = 0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

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