Question

Problem 7 [10] Twe point charges are separated by a distance of 10.0 cm on the x-axis. One has a charge of -25 C and the other +50 C. a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. b) If an electron (mass = 9.11 x10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)? Problem 8 [15] 4000 grams of water is placed in a vacuum chamber with dimensions of 2m by 2m by 2m at T=40°C. What is the final (ie. equilibrium) pressure inside the system? Assume the water vapor can be modeled as a ideal gas.

Answer 1

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Solution :- [a] 9,= -25 uc 2 = +5o uc d locm Р 2 cm E2 The direction of electric field (E) due to negative charge is attractive in nature. The direction of electric field (€₂) due to positive charge (92) is repulsive . Net Electic field (Enet ) € + E2 (Towards 9,) Ei URGO 22 Ez 1 1921 2 GAGO (d-r)2 E + Ez [12 1921 Putting 1911 YaGo [d-r]² all the value in above equation Enet = 8.987 25X106 2x102) = 8.987 (0.0625 +0.0078] ENET = 0.6318 V/m + Sox 106 (841022 • The direction of Electric field towards Negative Charge (9)

[b] kg mass of election (m) = 9.11 8 10-31 Charge of electron (e) = 1.681079 € Let the acceleration = a m/s2 Force (F) = ma We know that force in electric field (f = 9E) e E = ma -19 1.6X10 X 0.6318 m g.11 X10-31 = 11.09 x 1010 m/s2 2 Problem 8 [1s mass of water (m) = 4000 gram Molar mass of water (M) = 18 gram. no. of mole (n) = yooo m = 222 м 18 Dimension of vacum chamber = 2m x 2m Xam 3. .. Volume (v) - 8m 8000 litre Tempeature (T) = 40°c = 313.15 OK Pressure (P) = P atm universal Gros Constant (R) = 8.3144598 Now universal Gas Law equation PV = nRT DCOMINOTCO