จงเลือกคำตอบที่ถูกต้องท ่สุด ฟิสิกส์ ม.6

ฉนวน is the best material to generate electricity by friction because it is an insulator, meaning it does not conduct electricity well. When two objects made of different materials rub against each other, electrons can be transferred from one material to another, creating an electric charge. Insulators like ฉนวน do not allow this charge to easily flow through them, making them ideal for generating static electricity through friction.

The question provides information about the electric field strength and potential difference at a certain distance from a small spherical object. The electric field strength is given as 5 N/C and the potential difference is given as 20 V. Using the formula E = V/d, where E is the electric field strength, V is the potential difference, and d is the distance, we can rearrange the formula to solve for d. Plugging in the given values, we get 5 = 20/d. Solving for d, we find that the distance is 4 meters.

The correct answer is 4 -12ซม. The question asks for the distance at which a charge of 10 microcoulombs should be placed in order to have a net force of zero acting on it. According to Coulomb's Law, the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are 20 microcoulombs and 5 microcoulombs, and the distance is 6 cm. By setting up the equation and solving for the distance, we find that the distance should be 12 cm. Therefore, the correct answer is 4 -12ซม.

The distance between points A and B is given as 2 and 12 meters respectively. To move a charge of 4 Coulombs from point B to point A, we need to calculate the work done in kilojoules. The formula for work done is W = qV, where q is the charge and V is the potential difference. Here, the potential difference is the distance between the points, which is 12 - 2 = 10 meters. Plugging in the values, we get W = (4 x 10^-6) x 10 = 40 x 10^-6 = 40 x 10^-3 = 40 mJ. Converting this to kilojoules, we divide by 1000, giving us 40/1000 = 0.04 kJ. Therefore, the answer is 60.

The question states that point P is located 4 cm away from point A and that line AB is perpendicular to line PA at point P. The question also states that the electric potential at point P is twice the electric potential at point B. When we move the charge -q2 to point A, the ratio of q1 to q2 is 1.2.

When the metal block with a charge of 5 coulombs and a potential of 10 volts is touched to the other metal block with a capacitance of 1 farad and no initial charge, the charge on the first metal block will distribute itself between the two blocks. The final potential difference between the two blocks can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Plugging in the values, we get 5 = 1 * V, which gives V = 5 volts. Therefore, the potential difference on the first metal block will change to 5 volts, which is equivalent to 3.3 volts.

If the charge given to both conductors is the same, and the radius of conductor B is twice that of conductor A, then the electric potential on conductor A will be twice that of conductor B. Therefore, the electric potential on conductor A will be 2 times that of conductor B.

The distance between points A and B is given as 4 cm. Since point C is the point where the electric potential is zero, it must be equidistant from points A and B. Therefore, the distance AC is equal to half the distance AB, which is 2 cm. Hence, the correct answer is 3 cm.

The electric potential difference between two points is given by the formula V = kq/r, where V is the potential difference, k is the electrostatic constant, q is the charge, and r is the distance between the points. In this question, the charge +q is located at point C and the charge -q is located at point D. Points E and F are equidistant from the charge -q, and points C and D are equidistant from the charge +q. Therefore, the potential difference between points C and E, and between points D and F, will be the same. Hence, if a charge of +1 unit is moved between two points, the potential difference between points C and E and between points D and F will be the same.

The potential difference between the two conductors can be calculated using the formula V = Ed, where V is the potential difference, E is the electric field, and d is the distance between the conductors. In this case, the distance between the conductors is given as 2.0 cm or 2.0 x 10-2 m. Since the electric field is constant, the potential difference is also constant. Therefore, the potential difference between the conductors is 2.6 x 104 volts.

To calculate the potential difference, we can use the equation V = W/q, where V is the potential difference, W is the work done, and q is the charge. In this case, the work done is given as 4x10^-12 kg * 100 m/s = 4x10^-10 kg m^2/s, and the charge is 8x10^-9 C. Plugging these values into the equation, we get V = (4x10^-10 kg m^2/s) / (8x10^-9 C) = 0.05 V = 2.5 volts. Therefore, the potential difference required is 2.5 volts.

เมื่อมีความต่างศักย์ไฟฟ้า 100 โวลต์ และระยะระหว่างแผ่นขั้วโลกเท่ากับ 0.8 ซม., หยดน้ำมันที่มีมวล 8 x 10-16 กิโลกรัมจะได้รับพลังงานจากความต่างศักย์ไฟฟ้า 100 โวลต์ โดยใช้สูตรพลังงานไฟฟ้า = ความต่างศักย์ไฟฟ้า x ปริมาณชาร์จ電量 จะได้ พลังงานไฟฟ้า = 100 x (8 x 10-16) = 8 x 10-14 จูล์ จากนั้นใช้สูตรจำนวนอิเล็กตรอน = พลังงานไฟฟ้า / พลังงานอิเล็กตรอน จะได้ จำนวนอิเล็กตรอน = (8 x 10-14) / (1.6 x 10-19) = 5 x 105 ตัว ซึ่งเป็นจำนวนอิเล็กตรอนที่หยดน้ำมันได้รับ

If the force on charge -Q is equal in magnitude but opposite in direction to the force on charge P, it means that the charges are in equilibrium. This implies that the electric field at point C and point A is equal in magnitude but opposite in direction. Since the electric field is directly proportional to the charge, the magnitudes of the charges at point C and point A must be equal. Therefore, the magnitude of charge -Q is 7 x 10-3 C (since the magnitude of charge P is given as 26 x 10-3 C). However, since the question asks for the magnitude of charge -Q, the answer is -7 x 10-3 C.

The maximum electric charge that can be stored on a conductor is directly proportional to the product of the maximum voltage and the capacitance of the conductor. In this case, the maximum voltage is given as 9 x 105 volts and the conductor is a circular shape with a radius of 30 cm. The capacitance of a cylindrical conductor can be calculated using the formula C = 2πεr, where C is the capacitance, ε is the permittivity of free space, and r is the radius of the conductor. By substituting the given values into the formula, we can calculate the capacitance and then determine the maximum electric charge that can be stored. The correct answer is 30 microcoulombs.

The given question asks for the capacitance of a spherical metal with a radius of 10 cm in picofarads. The capacitance of a spherical capacitor is given by the formula C = 4πεr, where C is the capacitance, ε is the permittivity of free space, and r is the radius of the sphere. In this case, the radius is given as 10 cm. Plugging in the values, we get C = 4π(8.85x10^-12)(10) = 1.11x10^-9 F = 1.11 pF. Therefore, the correct answer is 11 pF.

The correct answer is "โลกมีศักย์ไฟฟ้าเป็นกลาง" because when an object has a positive or negative electric charge, it creates an electric field around it. This electric field interacts with the electric field of the Earth, which is neutral or has a balanced electric charge. By connecting the object to the ground, it allows the excess charge to flow into the Earth, equalizing the electric potential and making the object neutral. This is why it is necessary to connect the object to the ground to make its electric charge either positive or negative.

The given question describes a square ABCD with side length a. At angles A, B, and C, there are charges of +Q, +2Q, and Coulomb, respectively. At point E, which is a distance a away from angle D, the question asks how many Coulombs of charge are needed to make the resulting electrical force on charge +Q equal to zero. The correct answer is +8Q, meaning that a charge of +8Q is needed at point E to balance out the forces and make the net electrical force on charge +Q equal to zero.

If the inclined plane is not frictionless, the object B will experience a gravitational force down the incline and a frictional force opposing the motion. Since the object is at rest, the frictional force must be equal to the component of the gravitational force parallel to the incline. Using trigonometry, we can determine that the component of the gravitational force parallel to the incline is equal to the mass of object B multiplied by the acceleration due to gravity and multiplied by the sine of the angle of inclination. Setting this equal to the frictional force and solving for the height, we find that the height is equal to 0.303 meters.

The given particle has a mass of 2.0 x 10-5 kg and a charge of +2.0 x 10-3 C. When placed in an electric field that is directed vertically downward, the particle experiences an acceleration of 20 cm/s2. The magnitude and direction of the electric field can be determined using the equation F = qE, where F is the force experienced by the particle, q is the charge of the particle, and E is the electric field strength. Rearranging the equation to solve for E, we have E = F/q. Plugging in the values, we get E = (2.0 x 10-3 C) / (2.0 x 10-5 kg) = 100 N/C. Since the particle is experiencing an acceleration in the opposite direction to the electric field, the electric field must be directed downward. Therefore, the correct answer is 96 N/C with a downward direction.

The given statement states that if equal positive charges are brought close to the ends of the metallic rod, the distribution of charges on parts A, B, and C of the rod will be negative for A and C, but positive for B.