Quiz Listrik (Paket:03) Kelas 9 K13

1. Beda potensial antara dua buah titik pada kawat penghantar adalah 9 Volt. Berapakah energi listrik yang digunakan untuk memindahkan muatan sebesar 30 C diantara dua titik tersebut?

270 Joule

21 Joule

10/3 Joule

2,70 Joule

The potential difference between two points in a conductor is equal to the work done to move a charge between those points. The formula to calculate the electrical energy used is given by the equation: Energy = Potential difference x Charge. In this case, the potential difference is 9 volts and the charge is 30 coulombs. By substituting these values into the equation, we get: Energy = 9 volts x 30 coulombs = 270 Joules. Therefore, the correct answer is 270 Joules.

Explanation

The potential difference between two points in a conductor is equal to the work done to move a charge between those points. The formula to calculate the electrical energy used is given by the equation: Energy = Potential difference x Charge. In this case, the potential difference is 9 volts and the charge is 30 coulombs. By substituting these values into the equation, we get: Energy = 9 volts x 30 coulombs = 270 Joules. Therefore, the correct answer is 270 Joules.

2. Gambar berikut yang dapat menimbulkan gaya tolak terbesar, jika kedua muatan pada balon sama adalah….

A

B

C

D

not-available-via-ai

Explanation

not-available-via-ai

3. Perhatikan gambar berikut!Pernyataan yang benar untuk peristiwa gambar berikut adalah….

Sisir dapat menarik sobekan kertas kecil, karena sisir bermuatan listrik positif setelah diipakai menyisir rambut.

Sisir dapat menarik sobekan kertas kecil, karena sisir bermuatan listrik positif setelah digosok dengan kain wol.

The correct answer is that the comb will receive electrons from the wool, making it negatively charged. When the comb is brought near the torn pieces of paper, they will be attracted to the comb. This is because opposite charges attract each other, and the negatively charged comb will attract the positively charged torn pieces of paper.

Explanation

The correct answer is that the comb will receive electrons from the wool, making it negatively charged. When the comb is brought near the torn pieces of paper, they will be attracted to the comb. This is because opposite charges attract each other, and the negatively charged comb will attract the positively charged torn pieces of paper.

4. Berapakah beda potensial kutub-kutub baterai sebuah rangkaian jika baterai tersebut membutuhkan energi sebesar 60 J untuk memindahkan muatan sebesar 20 C?

120 Volt

12 Volt

3 Volt

1,20 Volt

The potential difference between the poles of a battery can be calculated using the formula V = W/Q, where V is the potential difference, W is the energy required, and Q is the charge moved. In this case, the energy required is 60 J and the charge moved is 20 C. Plugging these values into the formula, we get V = 60 J / 20 C = 3 Volt. Therefore, the correct answer is 3 Volt.

Explanation

The potential difference between the poles of a battery can be calculated using the formula V = W/Q, where V is the potential difference, W is the energy required, and Q is the charge moved. In this case, the energy required is 60 J and the charge moved is 20 C. Plugging these values into the formula, we get V = 60 J / 20 C = 3 Volt. Therefore, the correct answer is 3 Volt.

5. Sebuah baterai yang memiliki beda potensial sebesar 1,5. berapakah besar energi yang diperlukan baterai, jika memindahkan muatan sebanyak 75 C?

112,5 Joule

60 Joule

50 Joule

11,25 Joule

The energy required by the battery can be calculated using the formula: energy = potential difference * charge. In this case, the potential difference is given as 1.5 and the charge is given as 75 C. Therefore, the energy required by the battery is 1.5 * 75 = 112.5 Joule.

Explanation

The energy required by the battery can be calculated using the formula: energy = potential difference * charge. In this case, the potential difference is given as 1.5 and the charge is given as 75 C. Therefore, the energy required by the battery is 1.5 * 75 = 112.5 Joule.

6. Perhatikan gambar di samping!Berikut pernyataan yang benar tentang peristiwa yang dialami penggaris plastik setelah keduanya digosok kain wol adalah…

salah satu penggaris dengan tali sedemikian sehingga pada penggaris tersebut tidak ada gaya lain yang bekerja selain gaya gravitasi.

Ketika kamu menggosok-gosok mistar plastik dengan kain wol, kamu telah membuat kedua mistar plastik tersebut bermuatan negatif.

Ketika penggaris plastik digosok dengan kain wol, elektron dari kain wol dipindahkan ke penggaris plastik, membuat penggaris plastik bermuatan negatif. Ketika dua muatan negatif didekatkan, mereka akan saling menolak karena muatan yang sama. Oleh karena itu, penggaris plastik akan mengalami gaya penolakan atau gaya yang menentang untuk bersentuhan.

Explanation

Ketika penggaris plastik digosok dengan kain wol, elektron dari kain wol dipindahkan ke penggaris plastik, membuat penggaris plastik bermuatan negatif. Ketika dua muatan negatif didekatkan, mereka akan saling menolak karena muatan yang sama. Oleh karena itu, penggaris plastik akan mengalami gaya penolakan atau gaya yang menentang untuk bersentuhan.

7. Besar kuat medan listrik pada suatu titik yang berjarak r dari muatan Q dipengaruhi oleh besarnya gaya Coulomb, dan besarnya muatan listrik, jika jarak kedua muatan berada pada posisi 3 cm, dan kuat medan listrikyang dialami oleh muatan uji A terhadap muatan B sebesar 90 N/C, maka besar muatan B adalah….

90 x 10-5C

9 x 10-5C

9 x 106C

9 x 10-6C

The given answer of 90 x 10-5C is correct because it matches the given information that the electric field strength experienced by charge A due to charge B is 90 N/C. The electric field strength is given by the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge, and r is the distance between the charges. Rearranging the equation, we have Q = Er^2/k. Substituting the given values, we get Q = (90 N/C) * (0.03 m)^2 / (9 x 10^9 Nm^2/C^2) = 90 x 10-5C.

Explanation

The given answer of 90 x 10-5C is correct because it matches the given information that the electric field strength experienced by charge A due to charge B is 90 N/C. The electric field strength is given by the equation E = kQ/r^2, where k is the Coulomb's constant, Q is the charge, and r is the distance between the charges. Rearranging the equation, we have Q = Er^2/k. Substituting the given values, we get Q = (90 N/C) * (0.03 m)^2 / (9 x 10^9 Nm^2/C^2) = 90 x 10-5C.

8. Jika ada dua benda bermuatan listrik seperti 4 gambar berikut, maka benda manakah yang memiliki beda potensial yang lebih besar?

A

B

C

D

The object C has a larger potential difference because it is farther away from the reference point or the ground. The potential difference is directly proportional to the distance from the reference point. Therefore, the farther the object is from the reference point, the larger the potential difference it will have.

Explanation

The object C has a larger potential difference because it is farther away from the reference point or the ground. The potential difference is directly proportional to the distance from the reference point. Therefore, the farther the object is from the reference point, the larger the potential difference it will have.

9. Gaya Coulomb yang dialami kedua muatan A dan B adalah sebesar 3 x 10^-4N. Jika besar muatan A dan B masing-masing sebesar 2 x 10^-6C dan 1,5 x 10^-6C, berapakah besar kuat medan listrik yang dirasakan muatan A oleh muatan B?

150 N/C

15 N/C

1,5 x 10 N/C

0,15 x 102 N/C

The electric field strength experienced by a charge is equal to the force experienced by the charge divided by the magnitude of the charge. In this case, the force experienced by charge A is 3 x 10^-4 N and the magnitude of charge A is 2 x 10^-6 C. Therefore, the electric field strength experienced by charge A is (3 x 10^-4 N) / (2 x 10^-6 C) = 150 N/C.

Explanation

The electric field strength experienced by a charge is equal to the force experienced by the charge divided by the magnitude of the charge. In this case, the force experienced by charge A is 3 x 10^-4 N and the magnitude of charge A is 2 x 10^-6 C. Therefore, the electric field strength experienced by charge A is (3 x 10^-4 N) / (2 x 10^-6 C) = 150 N/C.

10. Agar mengetahui besar kuat medan listrik muatan Q, sebuah muatan uji positif (q_o) yang muatannya jauh lebih kecil diletakkan di dekat muatan tersebut dengan jarak r. Berdasarkan hukum Coulomb, muatan q_o tersebut akan mendapatkan gaya tolak dari muatan Q, karena kuat medan listrik (E) didefinisikan sebagai besarnya gaya listrik (F) yang bekerja pada satu satuan muatan uji (q_o), maka besarnya kuat medan listrik yang dialami oleh muatan uji, jika besar muatan bola A senilai 5 x 10^-9C dan muatan uji bola B senilai 5 x 10^-15 C dapat menghasilkan gaya coulomb senilai 10 x 10^-5N.

20 x 10-10N/C

20 x 1010N/C

2 x 10-10N/C

2 x 1010N/C

The correct answer is 2 x 10-10N/C. According to Coulomb's law, the electric field strength (E) experienced by a test charge is equal to the force (F) exerted on the test charge divided by the magnitude of the test charge (qo). In this case, the force is given as 10 x 10-5N and the test charge is 5 x 10-15C. Dividing the force by the test charge gives us the electric field strength of 2 x 10-10N/C.

Explanation

The correct answer is 2 x 10-10N/C. According to Coulomb's law, the electric field strength (E) experienced by a test charge is equal to the force (F) exerted on the test charge divided by the magnitude of the test charge (qo). In this case, the force is given as 10 x 10-5N and the test charge is 5 x 10-15C. Dividing the force by the test charge gives us the electric field strength of 2 x 10-10N/C.