Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
Answer:
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Explanation:
You are welcome.
Answer:
1) the capacitance of the capacitor increases. This is due to the induction of opposite charges on the two surfaces of the dielectric by the plate, this increased the charge in the field, from C =Q/v, it is seen that capacitance C will increase with increase in Q since v is constant.
2) the electric field intensity will also increase with increase in electric charges provided plate separation d remains constant.
