Answer:
bend toward the normal line
Explanation:
When light passes from a less dense to a more dense substance, (for example passing from air into water), the light is refracted (or bent) towards the normal. In your question the light is moving from rarer to denser medium
Answer:
Explanation:
The force exerted in a magnetic field is given as
F = q (v × B)
Where
F is the force entered
q is the charge
v is the velocity
B is the magnetic field
Given that,
The magnetic field is
B = 2•i + 4•j. T
The velocity of the electron is
v = 2•i + 6•j + 8•k. m/s
Also, the charge of an electron is
q = -1.602 × 10^-19 C.
Then note that,
V×B is the cross product of the speed and the magnetic field
Then,
F = q (V×B)
F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)
Note
i×i=j×j×k×k=0
i×j=k. j×i=-k
j×k=i. k×j=-i
k×i=j. i×k=-j
F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]
F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]
F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]
F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)
F = -1.602 × 10^-19(16•j - 32•i)
F = -1.602 × 10^-19 × ( -32•i + 16•j)
F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j
Then, the x component of the force is
Fx = 5.126 × 10^-18 N
Also, the y component of the force is
Fy = -2.563 × 10^-18 N
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
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